resultant of force

7/28/2019 Resultant of Force

1/27


2/27

It is a force or a couple that will have

the same effect to the body, both in

translation and rotation, if all the

forces are removed and replaced by

the resultant.


3/27

= = 1 + 2 + 3 +

= = 1 + 2 + 3 +

-The x-component of the resultant is equal to the

summation of forces in the x-direction.

-The y-component of the resultant is equal to the

summation of forces in the y-direction.


4/27

-The line of action of each forces in coplanar

concurrent force system are on the same plane.

All of these forces meet at a common point,

thus concurrent.


5/27


6/27

=

=

=

+

tan =


7/27

Three ropes are tied to a small metal ring. At the end of each rope

three students are pulling, each trying to move the ring in their direction.

If we look down from above, the forces and directions they are applying

are shown in Fig. 011. Find the net force on the ring due to the three

applied forces.


8/27

=

=30cos3750cos4580cos60

= 51.40 = 51.40

= =30sin37+50sin4580sin60 = 15.87

= 15.87


9/27

= + = 51.40 +15.87 = 53.79

tan= = 15.8751.40

=17.16

The net force on the ring is 53.79 lb

downward to the left at x = 17.16.


10/27

Find the resultant vector of

vectors A and B shown in Fig. P-012.


11/27

= 17 + 44 2 17 44 cos 70

= 41.39 /

sin70 =17

sin

s i n =17sin70

=17sin70

41.39 =22.70

=50+=50+22.70

=72.70


12/27

Three vectors A, B, and C are shown in the figure below. Find one

vector (magnitude and direction) that will have the same effect as

the three vectors shown in Fig. P-013 below.


13/27

=100cos37+100cos150+80cos233

= 54.88 = 54.88 =100sin37+100sin150+80sin233

= 46.29

= + = 54.88 +46.29

=71.79


14/27

tan=

tan= 46.2954.88

=40.15 =18040.15=139.85

= 71.79 139.85


15/27

Non-Concurrent Force System

The line of action of each forces in non-

concurrent force system lies on more than one

side of the plane.


16/27

Equations

Rx = Fx Ry = y

R =

+

tan =

MR = MORd= MR

Ryix= MR

Rxiy= MR


17/27

Where,

Fx = component of forces in the x-direction

Fy = component of forces in the y-direction

Rx = component of thew resultant in x-direction

Ry = component of thew resultant in y-direction

R = magnitude of the resultantx = angle made by a force from the x-axis

MO = moment of forces about any point O

d = moment arm

MR = moment at a point due to resultant force

ix = x-intercept of the resultant R

iy = y-intercept of the resultant R


18/27

Determine the resultant of the force

system and its x and y intercepts.


19/27

= = 30030 224 +361

= 149.895 = =30030+224 361

= 59.613 = + = 149.895 +59.613 = 161.314t a n = t a n = ..=21.69


20/27

The effect of a certain non-concurrent force system is

defined by the following data: Fx = +90 kN,Fy = -60

kN, and MO

= 360 kNm counterclockwise. Determine

the point at which the resultant intersects the x-axis.

MO = 360 kN-m

60a = 360

a= 6m

The x-intercept is

at 6 m to the left

of the origin.


21/27

In a certain non-concurrent force system it is found

that Fx = -80 lb, Fy = +160 lb, and MO = 480 lbft in a

counterclockwise sense. Determine the point at which

the resultant intersects the y-axis.

MO = 480 lb-ft

80b = 480

b= 6 ft.

The y-intercept of the

resultant is 6 ft above

the origin


22/27

Resultant of Parallel Force

The line of forcedoes not meet at

a common point.

Parallel forcescan be in the

same or in

opposite

direction


23/27

Equations

R= = 1 + 2 + 3 + Rd= = 11 + 22 +33 + Rectangular LoadR= woL

Triangular Load

R=

L

Trapezoidal Load

R= wo1L + 2 1


24/27

Where:

Fx = component of forces in the x-directionFy = component of forces in the y-direction

R = magnitude of the resultantW= weight of an object

L = length of an object


25/27

Determine the resultant of the four parallel forces

acting on the rocker arm.

R= R= 50 40 20 + 60

R = 50 lb downward

MO = Fd

MO = -50((6) + 40 (2)

20(3) + 60(8)

MO

= 200 lb-ft clockwise

Rd = MO

50d = 200

d= 4 ft to the right of O


26/27

A parallel force system acts on the lever shown.

Determine the magnitude and position of the resultant.

R=R= 30 + 60 - 20 + 40

R = 110 lb downward

MA = MA = 2(30)+5(60)-7(20)+11(40)

MA = 660 lb-ft counterclockwise

Rd= MA

110 d = 660

d = 6 ft to the right


27/27

The beam AB shown supports a load which varies an

intensity of 220 N/m to 890 N/m. Calculate the magnitude

and position of the resultant load.

F1 = 6(220)=1320 N

F2 = 6 670 = 2010N

R=F1 +F2=1320+2010

R = 3330 NRd = 3F1 + 4F2

3330d=3(1320)+4(2010)

d= 3.6 m

resultant of force

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