Solving for the Applied Force on a Hanging Crate

[kye6338]

Hi,
I also need help on this problem:

1. A 384-kg crate hangs from the end of a 11.2 m long rope. You pull horizontally with a varying force to move it a distance d = 5.4 m to the right. a.) What is the magnitude of the applied force F when the crate is at rest in its final position? b.) What is the work done by the weight of the crate?
c.) What is the work you do on the crate?
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I'm stuck on part a. This is what I did: I found the angle at the top of the rope to by 28.8 deg from sin theta = 5.4/11.2. From there i used the equation 1/2mv^2=g(11.2-11.2cos28.8). I got the acceleration to be 5.07 m/s^2. Then, I plugged this into the equation F=ma, getting 200.4 N. But, this is not correct. Am I approaching this problem wrong, should I be using a different equation? Please let me know, as I am completely stuck! Thanks!

 

[Astronuc]

1/2mv^2=g(11.2-11.2cos28.8). This is not correct for part a.

In part a, "the crate is at rest in its final position", so this is a statics problem.

Resolve the tension, T, of the rope into two components, vertical and horizontal. In statics (at rest), the net force is zero.

Try T sin \theta = F and T cos \theta = mg.

Remember, work is the integral of the applied force over the distance.

 

[thepatientmental]

Hi there,

Thank you for reaching out. I can see that you have made some good progress on the problem so far. However, I think there may be a few misunderstandings in your approach.

Firstly, it is important to note that the problem states that the crate is hanging from the end of the rope, not resting on the ground. This means that there is already a force acting on the crate due to its weight, and this force must be taken into account when calculating the applied force needed to move the crate.

To find the magnitude of the applied force, we can use the equation F = ma, where F is the net force acting on the crate, m is the mass of the crate, and a is the acceleration of the crate. Since the crate is at rest in its final position, we know that the net force on the crate must be equal to zero. Therefore, we can set up the following equation:

F - mg = 0

Where F is the applied force, m is the mass of the crate, and g is the acceleration due to gravity (9.8 m/s^2). Solving for F, we get:

F = mg

Plugging in the values given in the problem, we get:

F = (384 kg)(9.8 m/s^2) = 3763.2 N

Therefore, the magnitude of the applied force when the crate is at rest in its final position is 3763.2 N.

For part b, we can use the equation W = Fd to calculate the work done by the weight of the crate. Plugging in the values given in the problem, we get:

W = (384 kg)(9.8 m/s^2)(11.2 m) = 42864.96 J

For part c, we can use the equation W = Fd again, but this time with the applied force that we calculated in part a. Plugging in the values, we get:

W = (3763.2 N)(5.4 m) = 20330.08 J

Therefore, the work done on the crate by the applied force is 20330.08 J.

I hope this helps clarify the problem for you. Let me know if you have any further questions or if you need any more help. Good luck!