# A Man Raises A Platform

1. Sep 21, 2010

### bmoore509

1. The problem statement, all variables and given/known data
A 27.5 kg person stands on a 55 kg platform. He pulls on the rope that is attached to the platform via the frictionless lower-right pulley. He pulls the rope at an angle of 29◦ to the horizontal, as shown in the figure below. Assume: g = 9.8 m/s2 . Ignore friction. The platform remains level.

If he pulls the platform up at a steady rate, how much force is he pulling on the rope?

2. Relevant equations
Fnet=ma

3. The attempt at a solution
I don't know where to start. I don't understand what my free body diagram would look like. Would it be the weight of the person and platform combined, with Force up and w down? I'm just really confused.

2. Sep 21, 2010

### fizzynoob

well its not a complicated FBD, the man standing with the combination of the platform contribute a FG downwards, but the platform contributes a Normal forces to keep the system in the Y direction in equilibrium (before he pulls on the rope).

initially (not pulling on rope)

Fy = N - Fg = 0; therefor Fg = N

when pulling the rope, you break down your y component of your force

Fy = N - Fg + Fpsin(t) = ma; Fp is the force of you pulling on the rope, but you only want the y component.

and

Fx = Fpcos(t) = 0 ; but the guys trying to go up, not side to side; therefor its zero

hope that helps

#### Attached Files:

• ###### FBD.png
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Last edited: Sep 21, 2010
3. Sep 22, 2010

### bmoore509

All of that makes sense but then I get that:
F(sint)=82.5a

I don't have acceleration to find out what the force should be.

1.F = 269.5 N
2. F = 323.4 N
3. Cannot be determined.
4. F = 808.5 N
5. F = 404.25 N
6. F = 202.125 N
7. F = 539 N
8. F = 606.375 N
9. F = 485.1 N

4. Sep 22, 2010

### Thaakisfox

The point is that there is no acceleration as the man is pulling at a steady rate

5. Sep 22, 2010

### bmoore509

But wouldn't it be
Fy = Fpsin(t) = 0

Meaning F = 0?

But I know that's not right.

6. Sep 22, 2010

### bmoore509

So, if he's pulling himself up at a constant rate, acceleration is zero. But what do I do with that in the situation of Force? The force isn't zero because he is exerting a force.

7. Sep 22, 2010

### bmoore509

I'm lost as what to do from here, then:
F(sint)=82.5a

If I don't have acceleration, what can I do with this formula?

8. Sep 22, 2010

### fizzynoob

Forces in Y = -(M1+M2)g+Fsin(t) = 0; //M1 = mass of man, M2 = mass of plank

solve for F

9. Sep 22, 2010

### bmoore509

-(M1+M2)g+Fsin(t) = 0; //M1 = mass of man, M2 = mass of plank

Okay, so:
-(27.5+55)*(9.8)+Fsin(29)=0
-808.5+Fsin(29)=0
Fsin(29)=808.5
F=808.5/sin(29)
F=1667.664927 N