Lecture October 25, 2019

Exercise 1 Box 1 (𝑀" = 5𝑘𝑔) and Box 2 (𝑀' = 8𝑘𝑔) are connected by a string as shown. The pulley is massless and the surface is frictionless. If the system is released from rest, after 1 s, Box 2 has fallen a distance of: A) 15.68 m B) 3.01 m C) 4.9 m D) 6.03 m E) 7.84 m Done in class

1

2

Multiple Choice 1 Three blocks (A, B, C), each having the same mass M, are connected by strings as shown. Block C is pulled to the right by a force that causes the entire system to accelerate. Neglecting friction, the net force acting on block B is:

a) 0 b) c) d) e) ANSWER: C done in class

!F / 3

!F / 2 2

!F / 3

!F

Exercise 2 In the figure on the next page, Block 1 (M1 = 3 kg) rests on a frictionless surface, and Block 2 (M2 = 4 kg) rests on Block 1. The friction coefficients between Block 2 and Block 1 are:𝜇* = 0.1, 𝜇. = 0.2. A force F = 20 N is applied Horizontally, as shown. F = 20 N A) Assume that when the force is applied to Block 2 there is no slipping, and the two blocks accelerate together. Find the acceleration of the blocks. Answer: 𝑎 = 2.86 2

.3

B) Draw a FBD on Block 2, which includes the 4 forces acting on it. Calculate the normal force of on Block 2 due to Block 1, FN21, and the maximum static friction, 𝑓.,267. Using the magnitude of the acceleration, a, from part A, and applying Newton’s second law to the horizontal component, calculate the magnitude and direction of the friction force acting on Block 2. Is the assumption of no slipping in part A correct? Partial Answer: 𝑓.,267 = 7.84𝑁, 𝐹 − 𝑓 = 𝑀'𝑎 → 𝑓 = 8.56𝑁 > 𝑓.,267. Assumption is wrong. Blocks will slip. C) Find the acceleration of block 1 and 2. Partial Answer: Block 2: 𝑓* = 𝑀'𝑔𝜇* = 3.92𝑁; 𝐹 − 𝑓* = 𝑀'𝑎' → 𝑎' = 4.02 2

.3;

Block 1: 𝑓* = 𝑀"𝑎" → 𝑎" = 1.31 2.3

IMPORTANT POINTS: 1) 𝑓*is the friction between Block 1 and 2. Since Block 2 rests on Block 1, it only depends on the mass of Block 2. Kinetic friction 𝑓*on Block 2 is to the left since it opposes Block 2 sliding to the right of Block 1. The Kinetic friction on Block 2 is due to Block 1. Using Newton’s third law, Kinetic friction 𝑓*on Block 1 is due to Block 2, must be acting to the right. 2) Note that the acceleration of Block 2, 𝑎' > 𝑎".

M1 = 3 kg

M2 = 4 kg

Multiple-Choice 2 A 5.0-kg crate is resting on a horizontal plank. The coefficient of static friction is 0.50 and the coefficient of kinetic friction is 0.40. After one end of the plank is raised so the plank makes an angle of 30° with the horizontal, the force of friction is: A) 0N B) 17N C) 20 N D) 25N E) 49N ANSWER: B

Exercise 3 A box of textbooks of = 24.8 kg rests on a loading ramp that makes an angle with the horizontal. The coefficient of kinetic friction is = 0.25 and the coefficient of static friction is = 0.35. A) As the angle is increased, find the minimum angle at which the box starts to slip. B) At this angle, find the acceleration once the box has begun to move. C) At this angle, how fast will the box be moving after it has slid a distance 5.0 m along the loading ramp?

A. As the angle is increased, find the minimum angle at which the box starts to slip.

y-component Using first law x-component If the box doesn’t then the friction force depicted in the diagram is a static friction force

When the box begin to slip, the x-component of the weight (mg) equals the maximum static friction force .

; ;

B. At this angle, find the acceleration once the box has begun to move. When x-component of the weight overcomes the static friction force and the box accelerates down the incline. The friction force opposing the motion will be a kinetic friction force

. Using the above diagram and Newton’s second law:

; :

,

C. At this angle, how fast will the box be moving after it has slid a distance 5.0 m along the loading ramp? Use the kinetic equation

m θµk

µs θ

θ

θcosmgn =f fs ≤ µsn = µsmgcosθ

θµθ cossin mgmg s=

cosθsinθ

=1µs

tanθ = µs = 0.35 θ = 19.3!

!3.19=θ

θµµ cosmgnf skk ==

xk mamgmg =− θµθ cossin θµθ cossin gga kx −=

ax = 9.8m / s2( )sin19.3! − 0.25 9.8m / s2( )cos19.3! ax = 0.92m / s2

vx2 = v0x

2 + 2axx = 2axx

vx = 2axx = 2 0.92m / s2( ) 5m( ) = 3m / s

ax ax

+x

+y n

n f f

mg mg q

q

More Multiple choices You are driving your car along the road and hit a patch that is so icy that there is effectively no friction between your tires and the road. Which statement is most accurate? A) Your car will start to spin B) Nothing you do will have any effect; without any friction, you will continue to move in a straight line at a constant speed, as there is no net external force on you. C) If you hit the accelerator, you will get out of the icy patch faster. D) If you hit the brakes, you will eventually stop. E) If you twist the steering wheel, you can turn into the skid ANSWER: B The speed of a 4.0-N hockey puck, sliding across a level ice surface, decreases at the rate of 0.61 m/s2. The coefficient of kinetic friction between the puck and ice is: A) 0.062 B) 0.25 C) 0.41 D) 0.62 E) 1.2 ANSWER: A,

A 400-N block is dragged along a horizontal surface by an applied force �⃗� as shown. The coefficient of kinetic friction is uk = 0.4 and the block moves at constant velocity. The magnitude of �⃗� is closest to: A) 100 N B) 150 N C) 200 N D) 290 N E) 400 N

ANSWER: B, done in class

HINT: 1) The weight of box is mg = 400 N; 2) May be easiest to substitute the answers to see which one obeys Newton’s law. However, the systematic method of applying Newton’s Law will also give the correct answer.

Exercise 4 In the Figure below, 𝑀C = 5𝑘𝑔, 𝑀D = 3𝑘𝑔, 𝜇. = 0.7, 𝜇* = 0.15. A) Assume that the Boxes are moving left. Draw FBD on the boxes, and apply Newton’s law to each box, individually. Solve the equations to find the acceleration and the tension in the rope. B) Assume that the Boxes are moving right. Draw FBD on the boxes, and apply Newton’s law to each box, individually. Solve the equations to find the acceleration and the tension in the rope. F = 10 N Done in class

A B

Exercise 5 In the diagram, box 1(M1=2kg) lies on frictionless incline of 53.1°, and is moving up the incline with acceleration . Box 1 is connected by an ideal rope through a frictionless pulley to box 2 (M2=7kg), which rests on a 36.9° incline with friction. Block 2 is acted on by a horizontal force of magnitude FA = 5N. The

tension (T = ?) and kinetic coefficient ( ) are unknown.

a) Draw a free-body-diagram (FBD) of all forces on block 1 (M1), which includes the direction of its acceleration. Calculate the tension, T. a +y +x Since there is no friction on this side, only the T Horizontal is important x-component:

36.9°

b) Draw a free-body diagram (FBD) of all forces acting on block 2 (M2). Use

this to determine the magnitude and direction of the friction force ,

acting on block 2. Calculate the coefficient of kinetic friction, , between surfaces of block 2 and incline. See next page for solution.

a=1.04mi s−2

µk = ?

FN ,1

F1,xNet =T −Fg1,x =M1a→T =M1gsin36.9! +M1a

T =2kg×9.8mi s−2 ×0.8+2kg×1.04mi s−2 =17.76N

Fg1,x Fg1

fk ,2

µk

36.9° 53.1°

M2

T = ? T = ?

FA = 5N

µk = ?M1

No Friction

a = 1.04 m•s-2

!v

In the diagram, box 1(M1=2kg) lies on frictionless incline of 53.1°, and is moving up the incline with acceleration . Box 1 is connected by an ideal rope through a frictionless pulley to box 2 (M2=7kg), which rests on a 36.9° incline with friction. Block 2 is acted on by a horizontal force of magnitude FA = 5N. The tension (T = ?) and kinetic coefficient ( ) are unknown.

a) Draw a free-body-diagram (FBD) of all forces on block 1 (M1), which includes the direction of its acceleration. Calculate the tension, T.

a +y +x Since there is no friction on this side, only the T Horizontal is important x-component:

36.9° (2 points) (2 points)

b) Draw a free-body diagram (FBD) of all forces acting on block 2 (M2). Use this to determine the magnitude and direction of the friction force , acting on block 2.

Calculate the coefficient of kinetic friction, , between surfaces of block 2 and incline. In diagram below the forces and acceleration are dented by the solid black arrows. +y The dashed arrows denote the direction (+ or -) of the components of and . It is in this solution for clarity, not for marks. x-component a ,

Diagram (2 points)

T

36.9°

36.9° (2 points)

+x Y-component

a=1.04mi s−2

µk = ?

FN ,1

F1,xNet =T −Fg1,x =M1a→T =M1gsin36.9! +M1a

T =2kg×9.8mi s−2 ×0.8+2kg×1.04mi s−2 =17.76N

Fg1,x Fg1

fk ,2

µk

FA Fg

FN2 FNet ,x = Fg2,x −T −FA ,x − fk ,2 =M2a Fg2,x =M2gsin36.9! = 41.16N

FA ,x = FA cos36.9! = 4N

FA =5N fk ,2 fk ,2 = Fg2,x −T −FA ,x −M2a

FA ,y fk ,2 = 41.16N −17.76N −4N −7kg×1.04mi s−2 =12.12N

FA ,x Fg2,y

Fg2

Fg2,x FNet ,Y = FN2 −FA ,y −Fg2,y =0

36.9° 53.1°

M2

T = ? T = ?

FA = 5N

µk = ?M1

No Friction

a = 1.04 m•s-2

!v

Extra Problem: E1, E2 and E3 E1 Below block 1 (M1 = 5 kg) rests on a frictionless surface, and block 2 (M2 = 2 kg) rests on block 1. The friction coefficients between 2 and 1 are:𝜇* = 0.2, 𝜇. = 0.4. F = 7 N A) Find force of friction (magnitude and direction) on block 2, and on Block1. Why must they be in the opposite direction? What is the accelerations of the blocks Partial Answer: magnitude is 7.84N, 𝑎 = 12

.3

B) If F = 28N, find force of friction on block 2 and block 1. Find their accelerations. Partial Answer: Block 2, 3.92N, 𝑎' = 1.96 2

.3; Block 1, 3.92N, 𝑎" = 4.82 2

.3

E2 In the diagram below, Block A of mass mA = 2.0 kg is moving up a 36.9° incline with a speed of v0 = 5.0 m/s. It is attached to block B of mass mB = 4.0 kg, by a massless frictionless rope-pulley system. The coefficient of kinetic friction between Block A and the surface of the incline is µk = 0.2. Block A moves up 0.5 m to the top of incline. Assume block A accelerates up incline and B is accelerating downward as in diagram.

v0 = 5.0 m/s a BLOCK A a BLOCK B 36.9°

A) Draw a free-body diagram of block B, then use Newton’s second to write an equation that includes the tension of the rope, T and the acceleration, a. B) Draw a free body diagram on block A showing all the forces on A and the direction of its acceleration, a. Draw the x-y axes parallel and perpendicular to the incline. C) Use the diagram in part B) to find the normal force (perpendicular to incline) and the force of friction (parallel to incline) on block A. D) Using diagram from part B, determine the x-component (parallel to incline) of the net force, and use the second law to write an equation that include the tension of the rope, T and the acceleration, a. E) The two equations from part A) and D) are two equations with two unknown a and T. Solve them to obtain the tension T and acceleration a.

B

M1 = 5 kg

M2 = 2 kg

PARTIAL ANSWERS: A) ; C) , ; D)

; E) Subtract part D by part A to obtain , , T = 23 N. E3 Below Box A, MA = 2 kg, is on a floor with kinetic friction coefficient,𝜇* = 0.15, and static coefficient of friction,𝜇. = 0.25. Box A is connect to a frictionless pulley system to Box B, MB = 3 kg, resting on a 36.9°, with no friction. Box B, is connected by a frictionless pulley system, to Box C, MC = 10 kg, as shown.

T − 39.2N = − 4kg( )a FN = 15.68N fk =3.14NT −14.9N = 2.0kg( )a 24.3N =6kg×a a= 4.05mi s−2

Draw FBD of all forces on the boxes. Use Newton’s law to calculate the acceleration and tensions.

TAB TAB

TBC

TBC