The wheels, axle, and handles of a wheelbarrow weigh W = 65 N. The load chamber and its contents weigh WL = 538 N. The drawing shows these two forces in two different wheelbarrow designs. To support the wheelbarrow in equilibrium, the man’s hands apply a force to the handles that is directed vertically upward. Consider a rotational axis at the point where the tire contacts the ground, directed perpendicular to the plane of the paper. Find the magnitude of the man’s force for both designs.
there's 2 pictures of a guy with a wheelbarrow and there's 0.400 m on one side of the WL and then 0.200 m between WL and W and then 0.700 m between W and the man with the force (F). the other picture there's 0.600 m between the WL and W and there's 0.700 m between W and the man with the force (F). I've been working on this for an hour I really need help!
I just worked this out, I'm not sure if my figure is right.
Did you guys learn about moment or torque?
Here is my solution, let me know if the figure is correct with your description:
http://i61.tinypic.com/554h2s.jpg
DonHo,
Thank you! That was spot on and was correct. -A physic's student a year later
To find the magnitude of the force applied by the man's hands (F), we need to consider the equilibrium of forces acting on the wheelbarrow.
In both designs, the total clockwise moment (torque) about the rotational axis must be balanced by the total counterclockwise moment. The torque can be calculated by multiplying the force with the distance from the rotational axis.
Let's analyze each design:
1. Design with 0.400 m between WL and W and 0.200 m between W and F:
In this design, the clockwise torque is due to the weight of the load chamber and its contents (WL) acting at a distance of 0.400 m, and the weight of the wheels, axle, and handles (W) acting at a distance of 0.600 m. The counterclockwise torque is due to the force applied by the man's hands (F) acting at a distance of 0.700 m.
The equation for equilibrium is:
Clockwise Torque = Counterclockwise Torque
(WL * distance_WL) + (W * distance_W) = (F * distance_F)
Substituting the given values:
(538 N * 0.400 m) + (65 N * 0.600 m) = F * 0.700 m
Now solve for F:
(215.2 N) + (39 N) = F * 0.700 m
254.2 N = F * 0.700 m
F = 254.2 N / 0.700 m
F ≈ 363.14 N
Therefore, the magnitude of the force applied by the man's hands in the first design (0.400 m between WL and W, and 0.200 m between W and F) is approximately 363.14 N.
2. Design with 0.600 m between WL and W and 0.700 m between W and F:
In this design, the clockwise torque is due to the weight of the load chamber and its contents (WL) acting at a distance of 0.600 m, and the weight of the wheels, axle, and handles (W) acting at a distance of 0.600 m. The counterclockwise torque is due to the force applied by the man's hands (F) acting at a distance of 0.700 m.
Using the same equation for equilibrium:
(WL * distance_WL) + (W * distance_W) = (F * distance_F)
Substituting the given values:
(538 N * 0.600 m) + (65 N * 0.600 m) = F * 0.700 m
Now solve for F:
(322.8 N) + (39 N) = F * 0.700 m
361.8 N = F * 0.700 m
F = 361.8 N / 0.700 m
F ≈ 516.86 N
Therefore, the magnitude of the force applied by the man's hands in the second design (0.600 m between WL and W, and 0.700 m between W and F) is approximately 516.86 N.
To summarize:
- In the first design, the force applied by the man's hands is approximately 363.14 N.
- In the second design, the force applied by the man's hands is approximately 516.86 N.