Physics 1307 Practice Quiz 4
Problem I A window washer is standing on a scaffold supported by a vertical rope at each end. The scaffold weights 200 N and is 3.00 m long. What is the tension in each rope when a 700 N worker stands 1.00 m from one end? Problem II: A rope of negligible mass is wrapped around a 225 kg solid cylinder or radius 0.400 m. The cylinder is suspended several meters off the ground with its axis oriented horizontally, and turns on that axis without friction. (a) If a 75 kg man takes hold on the free end of the rope and falls under the force of gravity, what is his acceleration? (b) What is the angular acceleration of the cylinder? Solutions
1.We need first to draw a free body diagram for the scaffold including all the relevant forces:
T2
T1
1 m
W1
W2
1.5 m
Where W 1= 700 N is the weight of the worker and W 2 =200 N is the weight of the scaffold , notice that this force is applied on the geometrical center since we are assuming that the plank has a uniform mass distribution. Now let us write the two conditions for equilibrium: first the sum of all forces is equal to zero:

F =0
T 1T 2 −W 1 −W 2 =0
T 1 T 2 =900N
and second the sum of all torques also has to be also zero, where we measure the torques from T2
1

 =0
1 m 700 N 1.5 m200 N−3 mT 1=0
333 N =T 1
thus T 2 =567 N . 2. Again we need a drawing with the physical situation: (Looking the cylinder from the front ) R
mg
Now, let us write Newton's second law in the case of torques:

 =I 
1
2
−R mg= M R 
2
−2 mg
=
MR
2
−2×75 kg×9.8 m/ s
=16.33rad / s2
225 kg×0.4 m
with the angular acceleration of the cylinder we can find the acceleration of the man falling. Since they are both attached to the same string, we have:
a = R
a = 0.4 m×16.33 rad / s 2= 6.53 m / s2
2