How Is the Force Calculated in the Hanging Crate Problem?

[demonelite123]

A 230 kg crate hangs from the end of a rope of length L = 12.0m. You push horizontally on the crate with a varying force F to move it a distance d = 4.00m to the side.
a) What is the magnitude of F when the crate is in this final position?
During the crate's displacement, what are
b) the total work done on it?
c) the work done by the gravitational force on the crate,
and
d) the work done by the pull on the crate from the rope?

i can't get part a) at all. the only formulas i know are W = (1/2)m(v² - v₀²) and W = F * d. how do i find the force?

my books answer is 797 N and i have no idea how they got it.

 

[lanedance]

draw a free body diagram of the crate

if i understand the question correctly, the crate will now be off the ground & the rope inclined, you know the sides of the triangle, find the angle teh rope makes with the vertical

now balance all the forces, the vertical force will be equal & opposite to the gravtational force, use your trinagle to find the horizontal force, which will be balanced by the push

 

[kerimek]

I would approach this problem by first understanding the physical principles involved. The crate is hanging from a rope, which means it is experiencing two forces: the force of gravity pulling it down and the tension force of the rope pulling it up. When the crate is stationary, these two forces are balanced and the crate is in equilibrium.

To move the crate horizontally, we need to apply a force that is greater than the force of gravity pulling it down. This force is what we are looking for in part a). To find this force, we can use the formula F = ma, where F is the force, m is the mass, and a is the acceleration. In this case, the crate has a mass of 230 kg and is moving with a constant velocity, so the acceleration is 0. Therefore, the force we need to apply to move the crate horizontally is also 0.

However, we are told that we are pushing on the crate with a varying force. This means that the force we are applying is changing as the crate moves. To find the average force, we can use the formula F = Δp/Δt, where Δp is the change in momentum and Δt is the change in time. Since the crate is moving with a constant velocity, the change in momentum is 0. Therefore, the average force we are applying is also 0.

Now, let's look at the other parts of the problem. When the crate is being moved horizontally, the total work done on it is equal to the force applied multiplied by the distance it is moved. In this case, the force is 0, so the total work done on the crate is also 0.

The work done by the gravitational force on the crate is also 0 because the crate is not moving vertically. The gravitational force is only doing work when the crate is being lifted or lowered.

Finally, the work done by the pull on the crate from the rope can be calculated using the formula W = Fd, where F is the force and d is the distance. In this case, the distance is 4.00m and the force is the same as the force we calculated in part a), which is 0. Therefore, the work done by the pull on the crate from the rope is also 0.

In conclusion, while the problem may seem confusing at first, understanding the underlying principles can help us solve it. In this case, because the