Physics 1A
Week 4 Questions
Fall 1998
1. When you pedal your bicycle with your feet pressing directly downward on the pedals, maximum torque is
produced when the pedal sprocket arms are in the horizontal position, and practically no torque is produced
when they are in the vertical position. Explain.
2. For each system, tell whether or not it is in equilibrium. If it is, then what is the tension in the rope on the
right (i.e., where the arrow is)? If it isn't, then how could you change the system (by adding, removing, or
changing ropes) so that it will be in equilibrium? The weight of each block is 10 N.
3. The broom balances at its center of gravity. If you saw the broom in two parts through the center of gravity
and then weigh each part on a scale, which part will be heavier?
4. A boy and a girl are sitting on opposite sides of a seesaw. The seesaw is 2 m long. The boy has a mass of
20 kg, and the girl a mass of 30 kg. Where should the pivot point for the seesaw be placed so that the
seesaw is "balanced", i.e. neither side tends naturally to rise or fall?
5. What is the magnitude and direction of the force being exerted by the wall on the beam? Assume that the
beam is massless and that there is no friction in the pulley.
wall
60o
1 kg
beam
Physics 1A
Week 4 Solutions
Fall 1998
1. Torque = force  moment arm. When pedals vertical moment arm and torque small. When pedals
horizontal moment arm and torque large.
2.
3. CCW torque of handle balances CW torque of head end
Moment arm of handle end > head
Mass handle end < head end
4. The pivot point needs to be placed where the center of mass of the system is located. This is at X = (m1x1 +
m2x2)/(m1 + m2) = (20  0 + 30  2)/(20 +30) = 1.2 m from the boy, or 0.8 m from the girl. Another way
of solving this is to see that the torques exerted on the seesaw must add to zero. Taking the origin to be the
pivot point, the torques from the boy and girl must cancel. So m1gx1 = m2gx2, with x1 + x2 = 2m, resulting
in the same answer. Note the similarity between this and one of problems from last week's tutorial.
5. The beam is in equilibrium, so the net force on it must be zero. So the net horizontal force must be zero,
and the net vertical force must be zero. The vertical forces are the 10 N tension in the rope (the tension is 10
N because the mass is 1 kg and has a weight of 10 N) and the upward force of the wall. So the upward force
exerted by the wall must be 10 N. Also, the rope exerts a leftward force of 10 N on the beam, so the wall
must exert a 10 N rightward force on it. So the magnitude of the overall force is 14 N, and the direction
(relative to the horizontal) is given by arctan 10/10 = 45 degrees. Note that this is not the direction of the
beam itself.