Question

The figure below is a long stick with evenly spaced marks.The distance between two adjacent marks is 13.90 cm.

For each of the following forces or set of forces determine the requested torque. Use Point L as the axis of rotation. Use the sign convention that a counterclockwise (CCW) torque is considered positive and a clockwise (CW) torque is considered negative.

A force of 18.0 N is directed upward perpendicular to the stick at Point G. What is the torque from this force?

A force of 29.0 N is directed downward perpendicular to the stick at Point G. What is the torque from this force?

A force of 12.0 N is directed upward perpendicular to the stick at Point C. What is the torque from this force?

A force of 28.0 N is directed upward at angle 32.1

Answer 1

Torque T=r*F sin(theta).....

A)T=5*13.9-69.5 cm...
F=18 N.....stick moves in CCW.....

T=12.51 N.m.......

B)T=.695 *29=-20.155 N.m....
riotates in CW....


C)T=3*0.139*12 =-5.004 N.m...rotates in CW..


D)T=5*0.139*28*sin(90-32.1)= -16.48 N.m
rotates in CW
E)T=0...since angle theta =0..hence sin?=0....


F) Tnet =12.51-5.004 = 7.506 N.m

Answer 2

Use the formula T = r X F = |r||F| sin(\theta) , |r| is the magnitude of the moment arm of the force from point G, |F| is the magnitude of the force, and \theta is the angle (clockwise positive) between the moment arm and the force

i) T = |r||F| sin(\theta) = 5d . 18 sin (90) = 12.51 Nm (using right hand rule)

ii) T = |r||F| sin(\theta) = 5d . 29 sin(-90) = -20.155 Nm

iii) T = |r||F| sin(\theta) = 3d . 12 sin(-90) = -5.004 Nm

iv) T = |r||F| sin(\theta) = 5d . 28 sin(90-32.1) = 16.485

v) T = |r||F| sin(\theta) = 5d . 29 sin(0) = 0

vi) T = sum( |r||F| sin(\theta)) = 5d . 18 sin(90) + 3d . 12 sin(-90)= 7.506