Man connected to harness standing on plank

[Jtwa]

Homework Statement

A professor (m= 75 kg )is seated on a light harness connected to a rope and pulley system. The professor's feet touch a uniform plank (m_plank = 15kg) which is supported by a hinge at the wall. A bucket (M) is suspended from the right end of the plank.

m = 75kg
m_plank = 15kg
l = 2m
L = 4.9m

Find: Label all the forces on a FBD
Find: T if M=0
[/B]

Homework Equations

ΣM_hing = 0

The Attempt at a Solution

I took the moment about the hinge and looked at the man by himself. I think I'm drawing my FBD diagram incorrectly.

 

[BvU]

Hello JT,

From your drawing it's clear that the rope is attached to the end of the plank. That should be in the problem statement !

If prof is sitting in the harness, I expect one more force in his FBD. Otherwise I think you have most of it covered. F_ry is the only horizontal force I see, so it may well be 0...

 

[J Hann]

Have you tried writing the equations for the torques about the hinge end of the plank?
Also, you know that T is constant throughout the length of the rope.
This greatly would simplify the solution since you don't need to consider the forces
acting through the hinge.

 

[haruspex]

Also, you know that T is constant throughout the length of the rope.

Well, not quite. Since the rope is attached to the end of the plank, the section of rope below the plank will be at a different tension.

 

[BvU]

Yes, but the problem statement reads M = 0

 

[haruspex]

Yes, but the problem statement reads M = 0

Which post is that in response to? It does not appear to contradict my post #4.
Also, that M=0 condition is only for the final part of the question.

 

[Mister T]

$T$ is the tension in the section of rope above the plank. The section below will have a tension equal to $Mg$ which will generally be different than $T$, particularly if $M=0$.

 

[BvU]

You are absolutely right ! (Haru and T2)

 

[Jtwa]

Since the rope is connected to the man, why isn't T=(mass of man)*g

 

[BvU]

Part of his weight is taken by the harness !

 

[Jtwa]

I'm still having trouble understanding this question. Can someone draw a FBD ? Is Normal Force suppose to be on the FBD?

 

[haruspex]

Part of his weight is taken by the harness !

No, that's why it could have been mg. The reason that T is not mg is that not all of the weight is taken by the harness.

 

[haruspex]

I'm still having trouble understanding this question. Can someone draw a FBD ? Is Normal Force suppose to be on the FBD?

An FBD relates to a single rigid body. There are two appropriate FBDs here, one for the man and one for the plank.
Your diagram depicted both, which is ok except that where those two bodies interact you need to show the equal and opposite forces between them.

 

[BvU]

Part of his weight = not all of the weight

 

[haruspex]

Part of his weight = not all of the weight

It's a question of emphasis in English. If you write that the reason that T is not mg is that part of his weight is taken by the harness it implies that the part taken by the harness accounts for the discrepancy. If you mean that the part not taken by the harness accounts for it then you need to write "only part of his weight...".

 

[BvU]

Didn't want to give too much emphasis, just get the OP to think it over.

Non-native english typist... But then again, weren't you in Australia ?

 

[haruspex]

Didn't want to give too much emphasis, just get the OP to think it over.

Non-native english typist... But then again, weren't you in Australia ?

I am now.

 

[BvU]

Can't beat Oxford English