as physics unit 7 forces in equilibrium - animated science · mr powell 2009 index 7 forces in...

AS Physics Unit 7 Forces in Equilibrium

Mr D Powell

Mr Powell 2009 Index

Forces in Equilibrium 7

The addition of vectors by calculation or scale drawing. Calculations will be limited to two perpendicular vectors. The resolution of vectors into two components at right angles to each other; examples should include the components of forces along and perpendicular to an inclined plane. Conditions for equilibrium for two or three coplanar forces acting at a point; problems may be solved either by using resolved forces or by using a closed triangle. Moment of a force about a point defined as force × perpendicular distance from the point to the line of action of the force; torque. Couple of a pair of equal and opposite forces defined as force × perpendicular distance between the lines of action of the forces. The principle of moments and its applications in simple balanced situations. Centre of mass; calculations of the position of the centre of mass of a regular lamina are not expected.


Chapter Map


7.1 Vectors and Scalars

1. What is a vector quantity?

2. How do we represent vectors?

3. How do we add and resolve vectors? Representing a vector Addition of vectors using a scale diagram Addition of two perpendicular vectors using a calculator Resolving a vector into two perpendicular components Spec 3.2.1


A Definition....

Scalars Vectors

mass, temperature, time, length, speed, energy

displacement, force, velocity, acceleration, momentum

TASK: Make a note of each quantity, then look at the units and give a reason why they have been sorted as such..


Displacement and velocity

A runner completes one lap of an athletics track.

400 m

What is her final displacement?

If she ends up exactly where she started, her displacement from her starting position is zero.

What is her average velocity for the lap, and how does it compare to her average speed?

What distance has she run?


What did you decide?

Despite the trip moving at various speeds, because it ended up at the starting point, the average velocity was zero. This will always be true when the final displacement is zero.


Vector equations

An equation is a statement of complete equality. The left hand side must match the right hand side in both quantity and units. In a vector equation, the vectors on both sides of the equation must have equal magnitudes and directions. Take Newton’s second law, for example:

force = mass × acceleration

Force and acceleration are both vectors, so their directions will be equal. Mass is a scalar: it scales the right-hand side of the equation so that both quantities are equal.

Force is measured in newtons (N), mass in kilograms (kg), and acceleration in ms-2. The units on both sides must be equal, so 1 N = 1 kgms-2.


Displacement vectors

Harry and Sally are exploring the desert. They need to reach an oasis, but choose to take different routes.

Harry travels due north, then due east.

Sally simply travels in a straight line to the oasis.

When Harry met Sally at the oasis, they had travelled different distances. However, because they both reached the same destination from the same starting point, their overall displacements were the same.

N


Vector notation

A scalar quantity is often represented by a lower case letter, e.g. speed, v. A vector quantity can also be represented by a lower case letter, but it is written or printed in one of the following formats to differentiate it from the scalar equivalent:

The value of a vector can be written in magnitude and direction form:

e.g. v = (v, θ)

Or as a pair of values called components:

e.g. a = (8, 6) or x

y

8

6

8

6

a


Vector addition

Displacement vectors can always be added ‘nose to tail’ to find a total or resultant vector.

This can be done approximately by scale drawing:

It can also be done by calculation, breaking each vector down into perpendicular components first and then adding these together to find the components of the resultant:

c + d = 2

3

-2

2 + =

0

5

a

b

a + b 10

7 x

y


R = √ 32 + 42

Calculating a resultant

When adding two perpendicular vectors, it is often necessary to calculate the exact magnitude and direction of the resultant vector. This requires the use of Pythagoras’ theorem, and trigonometry.

For example, what is the resultant vector of a vertical displacement of 3 km and a horizontal displacement of 4 km?

direction:

= 5 km

tan θ = 4/3

= 53°

θ = tan-1(4/3) 4 km

3 km

R

θ = √ 25

magnitude:

R2 = 32 + 42


Just as it is possible to add two vectors together to get a resultant vector, it is very often useful to break a ‘diagonal’ vector into its perpendicular components.

Vector components

This makes it easier to describe the motion of an object, and to do any relevant calculations.

vertical component

horizontal component


Calculating components

A vector can be separated into perpendicular components given only its magnitude and its angle from one of the component axes. This requires the use of trigonometry.

For example, what are the horizontal and vertical components of a vector with a magnitude of 6 ms-1 and a direction of 60° from the horizontal?

6 ms-1

60°

x

y

cos60° = x / 6 x = 6 × cos60

= 3 ms-1

sin60° = y / 6 y = 6 × sin60

= 5.2 ms-1


Pythagoras & Adding Vectors

TASK: Now try out the same techniques for two more triangles;

1) opp = 2m, adj = 8m

2) opp = 1.5m, adj = 4m

Answers 1. = 14 , R =8.25m 2. = 20.6 , R =4.27m


Applications...

TASK: Now try out the same techniques for two more triangles;

1) What is R & if opp = 54N, adj = 22N

2) What is adj & if opp = 54ms-1, R =

85ms-1

Answers 1. = 67.83 , R =58.3m 2. =39.4 , R =65.6ms-1


Splitting a vector into components...

TASK: Put on your maths brain and try and explain using trig why Vx is related to cos and Vy is related to sin


Resolving in a context....

TASK: 1) A bird is flying at a speed of 15ms-1 at an angle of

36 to the horizontal. What is its vertical speed.

2) A javelin is thrown at a resultant force of 100N at a 45 angle. What are the horizontal and vertical forces on it.

Answers 1. Vy =8.8ms-1 2. Vy =70.7N Vx = 70.7N


Questions to try on your own....

Answers .....


1. How many radians in a circle?

2. How many radians in quarter of a circle?

3. Work out in degrees for a right-angled triangle if hyp = 5, opp = 4, adj = 3?

4. Work out opp if = 450 and the adj = 7cm?

5. Work out tan if opp = 53.0m, adj = 42.0m?

6. Work out opp if = /2 and the adj = 3cm?

7. Work out in radians for a right-angled triangle if hyp = 5, opp = 4, adj = 3?

8. Rearrange sin=opp/hyp to make the subject.

9. Work out sin (/2) =, sin (2) = , cos () = , cos(cos-1(/4)) = , tan(/4) =

10.Use formulae & numbers from your calculator to prove that;

More Trig Practice

adj

opptan


Vector Problems 1 Question 1

Both sides contribute as vectors so double it to If using only pythag

do simply 2 x 15N Sin20 = 10.26 = 10.3N

label upper part of triangle as a.

Hence a2 = (152+152-2x152Cos140)0.5 = 28.19N, then half for

triangle. To 14.1N. Then pythag (152-14.12)0.5 = 5.12N

(x2) = 10.23N

a c b



3) A mass of 20.0kg is hung from the midpoint P of a wire. Calculate the tension in each suspending wire in Newton’s. Assume g = 10ms-2. (Hint resolve…..)

Use the idea of point of equilibrium. The forces must balance vertical and

horizontal.

Weight balances the tension so for each wire

Use the 2T Cos70 = 200

T Cos70 = 100

T = 100 / Cos70

= 292.38N


Vector Problems 1 Question 5/6

5) Magnitude = sqrt(35002+ 2792) = 3511.1m = 3510m Direction = tan-1 (279/3500) = 4.557° = 4.6° south from vertical 6) Vertical motion: 21 sin (43°) = 14.32 ms-1 Horizontal motion: 21 cos (43°) =15.358 ms-1 S = 3 x 15.358 ms-1 = 46.1 ms-1

3500km

0.279km



Use the parallelogram method to resolve the forces acting on this object placed. (Hint employ both cosine & sine rule). (4 marks)



Use the parallelogram method to find the resultant acting on this object and angle. (Hint employ both cosine & sine rule). (4 marks)



3) Two forces of magnitude 10.0N and F Newton’s produce a resultant of magnitude 30.0N in the direction OA. Find the direction and magnitude of F. (2 marks) (Hint use Pythagoras)



The graph shows a part completed vector diagram. You task is to find out the vector R by mathematical analysis. The vectors A & B are shown in both coordinate and bearing formats. Show working for both a mathematical method (2 marks) and drawn out scaled method. (2 marks) Hint: this is not as hard as it might initially look!


Vector Problems 2 Question 4 - stage 1

Finding the components of vectors for vector addition involves forming a right triangle from each vector and using the standard triangle trigonometry. The vector sum can be found by combining these components and converting to polar form.

http://hyperphysics.phy-astr.gsu.edu/hbase/vect.html


Vector Problems 2 Question 4 – stage 2

After finding the components for the vectors A and B, and combining them to find the components of the resultant vector R, the result can be put in polar form by

http://hyperphysics.phy-astr.gsu.edu/hbase/vect.html


Vector Problems 2 - Question 5

5b

N

N

6) Forces of 60.0N and F Newton’s act at point O. Find the magnitude and direction of F if the resultant force is of magnitude 30.0N along OX (2 marks)


Vector Problems 2 - Question 7

Forces of 60.0N and F Newton’s act at point O. Find the magnitude and direction of F if the resultant force is of magnitude 30.0N along OX (2 marks)

SinC

c

SinA

a

c2 = a2 + b2 - 2abCosC


Using the Sine or Cosine Rule

30N In this problem we can solve it with other methods of trigonometry. Make a triangle of vectors where we are trying to find length c and angle A;

120

c = F

b = 30N

a = 60N

A

Cosine Rule c2 = a2 + b2 - 2abCosC c2 = 60N2 + 30N2 – 2x60Nx30N xCos120 c2 = 4500N2 +1800N2 c2 = 6300N c = 79.4N c = F = 79.4N

Sine Rule Use reciprocal version of above SinC / c = SinA / a (60N x Sin 120) / 79.4N) = SinA Sin-1 (60N x Sin 120) / 79.4N) =A A = 40.876 = 41

SinC

c

SinA

a


7) Simultaneous Equations (A*+)

If this problem is to be solved mathematically we must establish two formulae. One in the vertical and one in the horizontal; 30N

The key is knowing that the resultant force is 30N at 0º. Hence; Resolving Horizontally; 30N = FCos + 60NCos(360-120) 30N = FCos + 60N x -0.5 30N = FCos - 30N 60N = FCos Eq 1 Resolving Vertically; 0 = FSin + 60NSin(360-120) 0 = FSin + 60NSin(240) 0 = FSin + 60N x 3/2 0 = FSin -51.96N 51.96N = FSin Eq 2

So by dividing Eq 2 by Eq 1 FSin / FCos = 51.95N / 60N tan = 0.866 = 40.89 F = 60N / cos = 79.4N Or F = 51.96N / sin = 78.96N


Wires Example - Correct

Vertically

1.20N = T1cos30+ T2cos60

1.20N = 0.866T1+ 0.5T2 Eq 1

Horizontally

T1sin30 = T2sin60

0.5T1= 0.866T2

T1= 1.732T2 Eq 2

Sub 2 into 1;

1.20N = 0.866T1+ 0.5T2

1.20N = 0.866 x 1.732T2 +0.5T2

1.20N = 1.99T2

T2 = 0.60N

&

T1= 1.732T2

T1= 1.732 x 0.6N

T1= 1.04N

So by turning it on its side we can use the same formulae (which is confusing but correct)

Left side right side


WRONG!

Vertically

1.20N = T1sin30+ T2sin60

1.20N = 0.5T1+ 0.866T2 Eq 1

Horizontally

T1cos30 = T2cos60

0.866T1= 0.5T2

T1= 0.577T2 Eq 2

Sub 2 into 1;

1.20N = 0.866T1+ 0.5T2

1.20N = 0.866 x 0.577T2 +0.5T2

1.20N = 1T2

T2 = 1.2N

&

T1= 0.577T2

T1= 0.577x1.2N

T1= 0.69N


Vertically

1.20N = T1sin60+ T2sin30

1.20N = 0.866T1+ 0.5T2 Eq 1

Horizontally

T1cos60 = T2cos30

0.5T1= 0.866T2

T1= 1.732T2 Eq 2

Sub 2 into 1;

1.20N = 0.866T1+ 0.5T2

1.20N = 0.866 x 1.732T2 +0.5T2

1.20N = 1.99T2

T2 = 0.60N

&

T1= 1.732T2

T1= 1.732 x 0.6N

T1= 1.04N

If we take the outside angle instead of the inner angle we can do this and use the triangle;

(90-)

Alternative Can do by using (90- ) as angle;


7.2 Balanced Forces

1. Why do we have to consider the direction in which a force acts?

2. When do two (or more) forces have no overall effect on a point object?

3. What is the parallelogram of forces? Equilibrium of a point object Testing three forces in equilibrium Practical - Coplanar forces


Why are the objects not moving...


Real Setup…

T F

w=mg

X Y

mgF sin

This simple experiment tests the idea that the vertical component of a resultant force “F” can be balanced by vertical “mg” force. So changing “mg” changes F and thus we can use the example as y = mx+c


Example Practical Results…

y = 0.4905x + 9E-15 R² = 1

0.00

1.00

2.00

3.00

4.00

5.00

6.00

0.0 2.0 4.0 6.0 8.0 10.0 12.0

We

igh

t A

dd

ed

(N

)

Force on meter (N)

Weight +/-0.01N

Force on Newton meter +/-0.2N

mass +/-1g Weight +/-

0.01N Angle +/-1

deg

4.0 200 1.96 24 5.0 250 2.45 24 6.0 300 2.94 24 7.0 350 3.43 24 8.0 400 3.92 24 9.0 450 4.41 26

10.0 500 4.91 27 ave 24.71

Sin (ave) 0.42

In terms of forces Hyp = F opp = mg Hence... opp/hyp = adj

mg/F=sin(theta)

Fsin(theta) = mg F = x m = sin(theta) y = mg


Equilibrium Point of an Object


Example 1


Example 1 (with numbers)

A father pulls a child of weight 60.0N back with a force of 50.0N. a) What is the tension in each rope? b) What is the angle from vertical that the child is held at?

Answers… a) (60.02 +50.02)0.5 = 78.1N each rope 78.1N /2 = 39.1N b) tan = 50/60 = 0.833 = 39.8


Example 2

Hint: Draw your triangle 1,2,3….. W, S, F to make a similar triangle


Example 2 (with numbers)

A block of weight = 30.0N sits on a slope and is held by frictional forces of 10.0N a) What is the support force? b) What is the angle from vertical that the slope is at?

Answers… a) 10.02 +S2 = 30.02 (30.02 – 10.02)½ = S =28.3 b) tan = F/S = 10/28.3 = 0.353 = = 19.5

Hint: Draw your triangle 1,2,3….. W, S, F to make a similar triangle


Practical Extension...

Assemble the equipment shown below. Use some weights of different types and measure the angles shown.

The basic system will adjust in angle to whatever masses you add.

The tension in each string must equal the force exerted by W1 or W2. This enables the problem to be solved by resolving

Try it practically so get a range of angles & weights then prove the equilibrium.

http://www.walter-fendt.de/ph14e/equilibrium.htm





Example 3


Example 1...

Using the following figures the problem is very simple...

W1 = 5N W2 = 3N W3 = 6N 1 = 150 2 = 124

F1 F2 F1sin

F1cos

F2sin

F2cos

1. If you think that F1 & F2 must resolve and their components in H & V directions must balance with the weights to make the point P be in equilibrium.

2. Define the problem as shown below with two new angles and triangles. Make sure you use the correct angles!


Example 1...

Using the following figures the problem is very simple...

W1 = 5N W2 = 3N W3 = 6N 1 = 150 2 = 124 =60 = 34

F1 F2 F1sin

F1cos

F2sin

F2cos

5.25.2

coscos

0

667.133.4

sinsin

0

21

3321

FF

H

FWFF

V


Plenary Question

A 2kg chicken rests on point C on clothes line ACB as shown. What is the minimum breaking strength of the line to ensure the bird does not break the line?

F1 F2

F1sin30

F1cos 30

F2sin45

F2cos45

30 45

NFF

NFF

NFF

V

402

202

2

2

2045sin30sin

0

21

21

21

3

2

2

2

2

3

45cos30cos

0

21

21

21

FF

FF

FF

H


Plenary Question

A 2kg chicken rests as point C on clothes line ACB as shown. What is the minimum breaking strength of the line to ensure the bird does not break the line?

NFF 40221 3

221 FF

NF

NFF

NFF

FF

NFF

68.12732.2

64.34

2

4033

4023

2

3

2

402

2

22

22

21

21

NNF

FF

35.103

268.12

3

2

1

21

F1=10.35N F2= 12.68N

F1sin30

F1cos 30

F2sin45

F2cos45

30 45

So we have now worked out both the tensions in the line that are there from the weight of the chicken. This means that the line must be of a strength equal or better than F2 = 12.68N


Resolution of Forces – ICT Link Self Check

http://www.walter-fendt.de/ph14e/forceresol.htm






7.3 Principle of Moments

1. Under what conditions does a force produce a turning effect?

2. How can the turning effect of a given force be increased?

3. What is required to balance a force that produces a turning effect?

4. Why is the centre of mass an important idea?

Turning effects, The principle of moments, Centre of mass, Centre of mass tests, Calculating the weight of a metre rule (A03 ISA Practice)


Centres of mass and gravity

The centre of gravity of an object is a point where the entire weight of the object seems to act.

An alternative definition is that the centre of mass or centre of gravity of an object is the point through which a single force has no turning effect on the body.

In a uniform gravitational field the centre of mass is in the same place as the centre of gravity.

The centre of mass of an object is a point where the entire mass of the object seems to be concentrated.


Equilibrium

A body persists in equilibrium if no net force or moment acts on it. Forces and moments are balanced.

Newton’s first law states that a body persists in its state of rest or of uniform motion unless acted upon by an external unbalanced force.

Bodies in equilibrium are therefore bodies that are at rest or moving at constant velocity (uniform motion).

F1

F1

F2 F2

equilibrium


The principle of moments

The principle of moments states that (for a body in equilibrium):

total clockwise moments =

total anticlockwise moments

4 N 6 N

5 m d

4 × 5 = 6d

This principle can be used in calculations:

What is d?

20 = 6d

d = 20 / 6

d = 3.3 m


Human forearm

60 N

20 N

4 cm

16 cm

35 cm

F Taking moments about the elbow joint:

4F = (16 × 20) + (35 × 60)

4F = 2420

F = 605 N

The principle of moments can be used to find out the force, F, that the biceps need to apply to the forearm in order to carry a certain weight. When the weight is held static, the system is in equilibrium.

60 N

schematic diagram

weight of arm = 20 N


0.2 m

3 N

The uniform metre rule shown is in equilibrium, with its centre of gravity marked by the arrow ‘weight’. Find the weight of the metre rule.

0.3 m 0.5 m

W

total anticlockwise moments = total clockwise moments

3 × 0.2 = weight × 0.3

weight = 0.6 / 0.3

weight = 2 N

Finding the weight of a metre rule


Work out the mass then Density of a ruler...


Example Results + errors!

For a wooden ruler a density is quoted at 750kgm-3 d1 = (0.2 +/- 0.001)m +/- 0.5% d2 = (0.2 +/- 0.001)m +/- 0.5% m = (96+/1)g = (0.096 +/- 0.001)kg +/- 0.1% Thickness = (0.0061 +/- 0.00001)m +/- 0.16% Width = (0.028 +/- 0.00001)m +/- 0.04% Length = (1 +/- 0.001)m +/- 0.1%

m1x d1 = m2 x d2

Error in calculated mass = 1.1%

Density = m/v

Error in density = 1.1%+ 0.16% + 0.04% + 0.1% = 1.4%

Value for our ruler density...

(562 +/- 8) kgm-3

So this must be a different wood!


7.4 More on Moments

1. What can we say about the support force on a pivoted body?

2. When a body in equilibrium is supported at two places, how much force is exerted on each support?

3. What is meant by a couple? Support forces - Single support problems - Two support problems - Couple Problems

ISA preparation: practical task – practice in recording measurements and plotting a graph written task Section A and B – opportunities to discuss reliability and to analyse data and errors


Theory of Moments…


Practical Investigating the bridge crane Aims 1. To use simple measurements of distance and to apply the principle of moments to a

model bridge crane. 2. To relate practical measurements to straight line graph theory. 3. To consider measurement errors and how to reduce them. You require the following equipment: 1. two metre rulers 2. two spring balances calibrated in Newton's 3. a set square 4. slotted masses of weight that can be measured using a balance 5. thread 6. scissors 7. two stands and clamps 8. graph paper Safety Ensure the practical arrangement is stable and will not topple over.


Setup Advice....

Ensure that the spring balances are vertical and that the ruler is horizontal each time before you make their measurements. You may need to use a set square and a second metre ruler to do this. Use straight line graph theory and the general equation;

y = mx + c Theory.... Taking moments about the point where spring balance S2 supports the bar gives:

S1D = Wd2 + 0.5W0D

S1 = (W/D) d2 + 0.5W0

Y = S1 , x = d2 , m = W/D c = 0.5Wo

D


Setup Advice....

Ensure that the spring balances are vertical and that the ruler is horizontal each time before they make their measurements. You may need to use a set square and a second metre ruler to do this. Use straight line graph theory and the general equation;

y = mx + c It is recommended that you carry out calculations in SI units in order to avoid confusion.

Results The weight W should be equal to the gradient of the line D. The weight W0 should be equal to 2 the y-intercept.

W = (6N +/- 1.8) N Wo = (1.4 +/- 0.42)N

S1D = Wd2 + 0.5W0D S1 = Wd2/D + 0.5W0 Y = S1 , x = d2 , m = W/D c = 0.5Wo

D


Answers to Questions 1 (a) Use the second metre ruler to measure the vertical height (above the bench) of the horizontal ruler at each end. Use the set square against the bench (assumed horizontal) to check the second metre ruler is vertical. Adjust the clamps holding the horizontal ruler if necessary and recheck it is horizontal. Use the set square to check the spring balances are vertical. (b) The spread of the points about the line of best fit gives an indication of reliability. If the points are spread too much about the line of best fit, the results unreliable and the measurements should be repeated. 2) (a) Ensuring the ruler is horizontal and the spring balances are vertical each time the support forces S1 and S2 are to be measured. Precision of Newton meter. (b) Repeat each measurement at least twice for each position of the weight and calculate the mean value of each support force. Use more precise instrument with better scale. 3) (a) The spring balance does not read zero when it is unloaded. (b) The gradient would be the same. The y-intercept would be different.


Couples and torques

A couple is a pair of forces acting on a body that are of equal magnitude and opposite direction, acting parallel to one another, but not along the same line.

The torque of a couple is the rotation force or moment produced.

Forces acting in this way produce a turning force or moment.

F

F

The forces on this beam are a couple, producing a moment or torque, which will cause the beam to rotate.

d


The torque of a couple

torque of a couple = force × perpendicular distance between

lines of action of the forces

F

F

P

d

x d – x

A point P is chosen arbitrarily. Take moments about P.

total moment = Fx + F(d – x)

= Fx + Fd – Fx

= Fd

There is a formula specifically for finding the torque of a couple.


Couples Extras...

If two forces act about a hinge in opposite directions, there is an obvious turning

effect called a couple. The resulting linear force from a couple is zero.

The couple is given by the simple formula:

G = 2 Fs

This strange looking symbol, G, is “gamma”, a Greek capital letter „G‟. Couples

are measured in Newton metres (Nm)

The turning effect is often called the torque. It is a common measurement made

on motors and engines, alongside the power. Racing engines may be quite

powerful but not have a large amount of torque. This is why it would not make

sense for a racing car to be hitched to a caravan, any more so than trying to win a

Formula 1 race in a 4 x 4.


Summary Question…

Without a book or any notes in front of you.

1) Express in formula form, in a diagram and numerically and with text (if possible) the principles of “equilibrium” that you have just covered in terms of forces and moments for

A) One support problem B) Couple C) A two support problem

Pass this to another student and ask them what they think then hand in your work at the end for checking by your teacher….

NB: You will need to make up your own weights and distances


7.5 Stability

1. What is the difference between stable and unstable equilibrium?

2. When is a tilted object going to topple over?

3. Why is a vehicle more stable, the lower its centre of mass?


What happens here?


Equilibrium…

Stable Equilibrium – if an object is displaced it will return to that point Unstable Equilibrium – if it is displaced slightly it will not return to that point

Toppling…

If centre of mass goes over the pivot point an object will topple…


Moments

If we consider moments about the pivot point…. d = perpendicular distance from line of action of force F = force applied W = weight (acting through middle of base) b = base of object Clockwise Moment = Fd Anticlockwise Moment = W* b/2 Hence we have three scenarios for the relation of balance or unbalance…

F

d W

b 1. Fd < Wb/2 2. Fd = Wb/2 3. Fd > Wb/2

NB: try it out and see how it feels with a block


Moments

Imagine that this box had a base = 1.4m and the force was applied at distance = 1.8m. The weight of the box was 800N what is the force needed to tilt the box to equilibrium? This scenario means that… Fd = Wb/2 F = Wb/2d F = (800N * 1.4m) / (2* 1.8m) F = 311N

F

d W

b


Forces Triangles...

Resolving Forces Parallel to Slope

F = W x sin Resolving Perpendicular to Slope

Sx + Sy = W x cos

hyp x sin = opp

hyp x cos = adj

Use page 103 Q3 for practice


Example.. The lorry has a wheelbase of 1.8m and

centres of mass unloaded of 0.8m from the ground. What is the maximum angle it can drive at before toppling over….

d = 0.8m b/2 = 1.8/2m = 0.9 Hence opp/adj = tan tan-1 (0.9 / 0.8) = 48.3

Resolving on a Slope... (Harder A-B)

• When trying to work out the angle at which a lorry will start to topple we can use the idea of Resolving Forces Parallel to Slope

• However, it is hard to see to better to rotate then turn into a balance problem

Left = Right Up = down Now we can simply say that Friction = Wsin (opp) - Eq1 Sx + Sy = Wcos (adj) - Eq 2 Now divide Eq1 by q2 to F/s = tan

W

F = b/2

d

Page 103 Worked Example


7.6 Equilibrium Rules

1. What condition must apply to the forces on an object in equilibrium?

2. What condition must apply to the turning effects of the forces?

3. How can we apply these conditions to predict the forces acting on a body in equilibrium?

Free body force diagrams The triangle of forces The conditions for equilibrium of a body


Visuals…


Scale Diagrams….

Another useful tool in AS is to draw a scale diagram with ruler and compass. Then we can use the idea of a closed triangle to show equilibrium and resolve

an unknown force. Just try it out with two vectors and an angle. Then work out the third force by using simple trigonometry. The actual line

length should match your calculations


Summary Questions….

a) Moments about “X” balance….. 200N x 0.5 = Wo x 2m Wo = 50N b) Support Forces must match the

total weight down… Sx = W0 + W = 50N + 200N = 250N

5m

0.5m 0.5m

W= 200N

W0= ?

Basic



a) Moments about “X”

The moment is F x d = 1.2m x 1500N = 1800Nm b) Hence, extra force is 1800N (as applied from 1m away!) 1.2m x 1500N = 1800N x ?

W= 1500N

Basic



1. Split the problem into two triangles working out the angle from vertical for each

2. Label up sides with adj1 = adj2 Resolve Left = Right

T1sin40 = T2 sin 30 Resolve vertical direction….

2.8N = T1cos40 + T2cos30

Now solve with simultaneous equations T1 = 1.493N = 1.5N T 2 = 1.911N = 1.9N

40 30

opp1 opp2

hyp2 hyp1 adj

Medium Physics/ Harder Maths



a) Weight acts down which makes it more complex!

• But if you turn problem to flat then work out

the component of W acting in line with T to balance…

• So green triangle you need to work out the Hyp using as same angle as slope

Angle of Beam to Horizontal opp = sin x hyp -> Sin-1(6m/10m) = 36.9 Up = Down (moments from “X”) T x 10m = 10m/2 x 15,000N cos (36.9) = 6000N

T

W = 15kN

W = 15kN

Harder Physics/ Harder Maths



Angle slope = 36.9 b) Summing forces in the…. x direction on the base of the girder… Sx (support force) = T*sin(36.9) Sx = T*sin(36.9) = 6000*sin(36.9) = 3602N = 3.6kN

Y-direction… Sy + 6000*cos(36.87) = 15,000 Sy = 15000 - 6000*cos(36.9) = 10202N = 10.2kN Resolve Resultant for Support Force…

R = (3.6kN2 + 10.2kN2) ½ = 10.8kN

T=6000N

3.6kN

10.2kN

Harder Challenge/ Harder Maths

Tsin


Quick Test what does each one represent... 1. A point on an object where the entire weight of the object seems to act. In a uniform

gravitational field, it is in the same place as the centre of mass.

2. A point on an object where the entire mass of the object seems to be concentrated. In a uniform gravitational field, it is in the same place as the centre of gravity.

3. A pair of forces acting on a body that are of equal magnitude and opposite direction, acting parallel to one another, but not along the same line.

4. The state a body is in if no net force or moment acts on the body.

5. The point about which a lever turns. Also called the pivot.

6. The turning effect of a force or forces. Can also be called the torque (symbol τ). It is calculated by multiplying the force by the perpendicular distance between the pivot and the line of action of the force. The units of moment are newton metres (Nm).

7. The point about which a lever turns. Also called the fulcrum.

8. The principle stating that the sum of the clockwise moments are equal to the sum of the anticlockwise moments acting on a body.

9. The turning effect of a force or forces, given the symbol τ. Can also be called the moment. It is calculated by multiplying the force by the perpendicular distance between the pivot and the line of action of the force. The units of torque are newton metres (Nm).

10. The rotation force or moment produced by a couple. It can be calculated by multiplying the force by the perpendicular distance between the lines of action of the forces. Its units are newton metres (Nm).

11. The force created by the gravitational attraction on a mass. Its units are newtons (N).


Quick Test Answers... 1. centre of gravity – A point on an object where the entire weight of the object

seems to act. In a uniform gravitational field, it is in the same place as the centre of mass.

2. centre of mass – A point on an object where the entire mass of the object seems to be concentrated. In a uniform gravitational field, it is in the same place as the centre of gravity.

3. couple – A pair of forces acting on a body that are of equal magnitude and opposite direction, acting parallel to one another, but not along the same line.

4. equilibrium – The state a body is in if no net force or moment acts on the body. 5. fulcrum – The point about which a lever turns. Also called the pivot. 6. moment – The turning effect of a force or forces. Can also be called the torque

(symbol τ). It is calculated by multiplying the force by the perpendicular distance between the pivot and the line of action of the force. The units of moment are newton metres (Nm).

7. pivot – The point about which a lever turns. Also called the fulcrum. 8. principle of moments – The principle stating that the sum of the clockwise

moments are equal to the sum of the anticlockwise moments acting on a body. 9. torque – The turning effect of a force or forces, given the symbol τ. Can also be

called the moment. It is calculated by multiplying the force by the perpendicular distance between the pivot and the line of action of the force. The units of torque are newton metres (Nm).

10. torque of a couple – The rotation force or moment produced by a couple. It can be calculated by multiplying the force by the perpendicular distance between the lines of action of the forces. Its units are newton metres (Nm).

11. weight – The force created by the gravitational attraction on a mass. Its units are newtons (N).


Exam Question ...Jan 07 Spec A Q5

5 (a) Define the moment of a force. (b) The diagram shows the force, F, acting on a bicycle pedal. (i) The moment of the force about O is 46Nm in the position shown. Calculate the value of the force, F. (2 marks) (ii) Force, F, is constant in magnitude and direction while the pedal is moving downwards. State and explain how the moment of F changes as the pedal moves through 80°, from the position shown. (2 marks)


Exam Question ...Jan 07 Spec A Q5

5b)…. (i) The moment of the force about O is 46Nm in the position shown. Calculate the value of the force, F. (2 marks) (ii) Force, F, is constant in magnitude and direction while the pedal is moving downwards. State and explain how the moment of F changes as the pedal moves through 80°, from the position shown. (2 marks)

46Nm

240N

as physics unit 7 forces in equilibrium - animated science · mr powell 2009 index 7 forces in...

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