Lesson 18 Homework Solutions
1.
A 30kg child sits on a seesaw 1.5 m from the pivot point. How far
from the pivot point on the other side will his 45kg friend has to sit
for the seesaw to be in equilibrium?
A is the pivot point. The
weight of the 30kg child is
294 N at a distance 1.5 m
from the pivot. The weight
of the 45kg child is 441 N
at a distance x from the
pivot.
294 N
A
1.5 m
x
441 N
The torque of 294 N about the pivot point A is:
τ
1
=  1.5 m
×
294 N =  441 mN (This is a clockwise torque)
The torque of 441 N about the pivot point A is:
τ
2
= x m
×
441 N = 441x mN (This is a counter clockwise torque)
When the seesaw is in equilibrium, the sum of the torques about A is
zero.
441 x – 441 = 0
x = 1 m
2.
A force of 36 N is applied to a particle located 0.22 m from its axis of
rotation. What is the magnitude of the torque about this axis if the
angle between the direction of the force and the torque arm is (i) 60
o
,
(ii) 90
o
?
r = 0.22 m, F = 36 N
(i)
θ
= 60
o
The component of the force at
right angles to the torque arm is
36 sin60
o
.
Torque about the pivot is:
τ
= r
×
F sin
θ
= 0.22
×
36 sin 60
o
= 6.86 m N
0.22 m
60
o
36 N
36 sin 60
o
(ii)
θ
= 90
o
τ
= r
×
F sin
θ
= 0.22
×
36 sin 90
o
= 7.92 m N
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document3.
A uniform meter ruler is pivoted at its center with a 150 g mass
suspended from the 30 cm mark. (i) At what position should a 100 g
mass be placed to keep the system in equilibrium? (ii) What mass
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '10
 George
 Physics, Force, Work, 10kg, 2.0 m, 12kg, 6.86 m, 7.92 m

Click to edit the document details