HELP: Normal Force/Tension/Work Problem

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In summary, the conversation is about a problem given by a teacher in which a block of cheese is on an elevator cab being pulled upward by a cable. The first part of the solution involves calculating the work done on the cab by the force from the cable, while the second part involves finding the magnitude of the normal force on the cheese. Some confusion arises and a fellow student asks for clarification. The solution is explained, with the conclusion that the acceleration of the cheese and elevator cab is zero due to the frame of reference. The final answer for the normal force is 2.45N when using the elevator cab as the frame of reference, and 4.902N when using no frame of reference.
  • #1
ctpengage
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Hey there guys.

Our teacher gave us an example problem to which he provided the solution for us to study.

The question is:
A 0.250 kg block of cheese lies on the floor of a 900 kg elevator cab that is being pulled upward by a cable through distance d1 = 2.40 m and then through distance d2 = 10.5 m. (a) Through d1, if the normal force on the block from the floor has constant magnitude FN = 3.00 N, how much work is done on the cab by the force from the cable? (b) Through d2, if the work done on the cab by the (constant) force from the cable is 92.61 kJ, what is the magnitude of FN?

The solution is:
(a) The net upward force is given by
Ftension + FN -(m + M)g = (m + M)a
where m = 0.250 kg is the mass of the cheese, M = 900 kg is the mass of the elevator cab,
Ftension is the force from the cable, and FN is the normal force on the cheese. On the cheese alone, we have
FN - mg = ma
Reaaranging and subbing in known values yields the acceleration as 2.20m/s2
Thus the force from the cable is
Ftension= (m+M) (a+g) − FN =10800N and the work done is therefore
10800N x (2.4m) = 25900J

I understand part 1 but I don't understand the solution to part 2.

Solution Part 2:
If W = 92.61 kJ and d2 =10.5 m , the magnitude of the normal force is

FN = (m+M)g - W/d2 = 2.45N

I don't understand the solution to part two at all. It seems that for part two, the acceleration of the cheese and elevator cab is zero. Why?

Can anyone help clarify issues with this problem.

Thanks
 
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  • #2


For guys who think I created some sort of fake scenario above to get you guys to help me cheat or something, the problem is no 25, chapter 7 from Fundamentals of Physics 8ed Extended and is a tutoring problem whose solution our prof. made available on WileyPLUS.
 
  • #3
please guys help ere :D:D:D
 
  • #4
We just got to this problem in my AP physics course. I thought the same thing as you when I arrived at the answer and checked it with the back of the book. The back of the book says that the normal force is equal to the weight of the cheese. The only way that is possible is if the accelerate or mass is zero. Since the mass is obviously not zero, it must be the acceleration. It then dawned on me that it must be because the acceleration of the cheese with the elevator cab as the frame of reference is zero. So then Fn= -mg = 2.45N. If you used no frame of reference the answer you should have arrived at is 4.902N.
 
  • #5


Hi there,

It seems like you are having trouble understanding the solution to part 2 of this problem. Let me try to clarify it for you.

In part 2, we are given the work done on the cab by the force from the cable (92.61 kJ) and the distance it is pulled (10.5 m). We are asked to find the magnitude of the normal force on the cheese.

To solve this, we need to use the same equation as in part 1, but rearranged to solve for the normal force (FN). The equation is:

Ftension + FN - (m+M)g = (m+M)a

Since we are given the work done (W) and distance (d2), we can use the formula for work, W = Fd, to substitute for the tension force (Ftension):

Ftension = W/d2

Substituting this into the original equation and solving for FN, we get:

FN = (m+M)g - W/d2

Plugging in the given values for mass (m+M), acceleration (g), and work (W), we get:

FN = (0.250 kg + 900 kg)(9.8 m/s^2) - (92.61 kJ)/(10.5 m)

Solving this, we get FN = 2.45 N.

To answer your question about why the acceleration is zero, it is because in this part of the problem, the cheese and the elevator cab are not moving. So, the acceleration is zero and the equation simplifies to just:

Ftension + FN = (m+M)g

I hope this helps clarify the solution for you. If you have any further questions, feel free to ask. Good luck with your studies!
 

1. What is the normal force?

The normal force is the force that a surface exerts on an object that is in contact with it. It is always perpendicular to the surface and acts in the opposite direction of the force exerted by the object on the surface.

2. How is normal force calculated?

Normal force is typically calculated using the formula FN = mgcosθ, where m is the mass of the object, g is the acceleration due to gravity, and θ is the angle between the surface and the direction of gravity.

3. What is tension?

Tension is the force that is transmitted through a string, rope, cable, or any other type of flexible material when it is pulled at both ends. It is directed along the length of the material and its magnitude is equal at both ends.

4. How is tension calculated?

Tension can be calculated using the formula T = F/A, where T is the tension, F is the force applied, and A is the cross-sectional area of the material.

5. What is work?

Work is the measure of energy transfer that occurs when a force is applied to an object and it is displaced in the direction of the force. It is calculated by multiplying the force applied by the distance over which the force is exerted.

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