Newton's Laws Homework question involving a pulley.

  • #1

Homework Statement



John (mass 80) rests on top of a light cube (mass = 0). A light rope passes horizontally from him over the front edge, and vertically down to Brian (mass 60kg) who hangs in contact with the front face as shown. The cube is on a horizontal surface and there is no friction anywhere. Joe (mas 70kg) has his back to a wall and applies a horizontal push perpendicular to the rear face. Calculate the value of the force, which will prevent both John and Brian from sliding over the surface of the cube.

Here is the diagram which i redrew from the book.

xfz4th.jpg


Homework Equations



Fun=ma

The Attempt at a Solution



I first found how fast the people would accelerate on top of the cube.

For the 80kg person: Ft=80a
For the 60kg person: -Ft+60(9.8)=60a
Add the two equations to solve for acceleration: 588=140a
So the acceleration for both of the is 4.2m/s^2

I think that means that John will have to accelerate the cube in a way such that it cancels out the acceleration of the people on top.

The combined weight of the people on the cube is 140kg, and they are accelerating at 4.2m/s^2.

I did:
Fun=ma
Fa(by john to the cube)=140(4.2)
Fa=588N.
So therefor, John must apply a force of 588N to prevent them from falling off.



I don't know if i got it right, or if i made the right assumptions. Does my answer look correct to you guys? Any hints if i got it wrong?

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
I drew a better diagram, because i now know what i did is wrong.

16kqkpl.jpg


The acceleration is 7.35m/s^2.

I do not know where to go from here. I know Joe has to apply a force on the cube. I know that Brian (the hanging one) will accelerate in the same direction as the applied force. I don't know how to figure out how big this force has to be to prevent John and Brian from slipping off the cube.

Can anyone help me out, now that i have the correct acceleration?
 
Last edited:
  • #3
1,198
5
The 7.35 m/sec^2 looks good because that puts enough tension in the cord to cancel the force of Brian.

To achieve that acceleration, the person standing on the ground would have to accelerate both Brian AND John at that rate. So what force would be needed if the block weighs nothing?
 
  • #4
The 7.35 m/sec^2 looks good because that puts enough tension in the cord to cancel the force of Brian.

To achieve that acceleration, the person standing on the ground would have to accelerate both Brian AND John at that rate. So what force would be needed if the block weighs nothing?
That would be 1029N.
Fun=140(7.35)

That looks too simple. Our teacher said this is a very difficult problem...
 
  • #5
1,198
5
Is your teacher a sarcastic one?

In my younger years before becoming an engineer I taught math and physics in high school. Once in a while I'd place an asterisk by the problem number of a problem that was no more difficult than any other. By far, more students would work the problem wrong because the asterisk psyched them out. I quit doing it.

Well if this problem is very difficult, then I'm missing the boat also. Got to run...
 
  • #6
Is your teacher a sarcastic one?

In my younger years before becoming an engineer I taught math and physics in high school. Once in a while I'd place an asterisk by the problem number of a problem that was no more difficult than any other. By far, more students would work the problem wrong because the asterisk psyched them out. I quit doing it.

Well if this problem is very difficult, then I'm missing the boat also. Got to run...
Yeah, he is. I just don't like how he said "Be very careful when you draw the free body for the whole box". He made it out to be crazy hard.
 

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