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This question was created from Test 3.8


Please answer question 14 and shows the steps for the answer please. Thank you


Version 013 -TEST03 - tsoi - (91875)
The pivot provides an external force to the
rod+clay system.
III) Angular momentum is conserved, since
there is no external torque applied to the
2.4 m
rod+clay system.
013 10.0 points
A horizontal bar with a mass m suspended
from one end is held by a cord with tension
T, fastened at a distance D from the end
of the bar that is attached to a wall with a
1.2 m
frictionless pin.
17.8 kg
Tension T
What is the minimum angle min (between
the horizontal and the ladder) so that the
person can reach a distance of 1.2 m without
having the ladder slip? The acceleration of
gravity is 9.8 m/s.
What is the torque about the pin exerted
1. 50.9925
on the bar by the cord?
2. 69.91
3. 61.8868
1. DT (1 - cos 0)
4. 51.9112
5. 63.4349
2. DT cos 0 correct
6. 59.8863
7. 59.0687
3. DT sin 0
8. 44.3186
9. 66.7323
4. DT (1 - tan 0)
10. 56.1397
5. DT (1 - sin 0)
Correct answer: 59.8863.
6. DT tan 0
The distance from the pin to the cord is
D cos 0, so the torque is
T X D cos 0 = DT cos0.
014 10.0 points
A 17.8 kg person climbs up a uniform ladder
with negligible mass. The upper end of the
ladder rests on a frictionless wall. The bottom
of the ladder rests on a floor with a rough
Let : g =9.8 m/s2 ,
surface where the coefficient of static friction
is 0.29. The angle between the horizontal and
d = 1.2 m,
the ladder is 0 . The person wants to climb up
L = 2.4 m ,
the ladder a distance of 1.2 m along the ladder
m = 17.8 kg, and
from the ladder's foot.
H = 0.29 .

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