Solving Weight Distribution for Iron Bar: 2 Man Problem

[konichiwa2x]

An iron bar of weight W is having its one end on the ground and the other on the shoulder of a man. The rod makes an angle alpha with the horizontal. What is the weight experienced by the man?

If the same iron bar were balanced by two men, what will be the be the weight of the bar as experienced by one of the men at the instant the other man let's go of the bar?


Can someone please give me an idea of how to do these two problems? thanks for your time.

 

[Hootenanny]

Think moments.

 

[konichiwa2x]

This question was under the chapter "Laws of motion". Principle of moments hasnt been taught to us yet.

 

[OlderDan]

This question was under the chapter "Laws of motion". Principle of moments hasnt been taught to us yet.

Does "Laws of motion" include angular motion? Perhaps you have encountered the term torque rather than the term moment.

 

[konichiwa2x]

I know torque = Force x distance
also torque = moment of intertia . angular acceleration.

but I still don't understand how to do this problem using this.

 

[OlderDan]

I know torque = Force x distance
also torque = moment of intertia . angular acceleration.

but I still don't understand how to do this problem using this.

For the first part, the bar is at rest. The net force acting on the bar must be zero and the sum of the torques must be zero. You treat the bar as if all the weight was acting at its center of gravity (the middle of the bar). You can calculate the torques about any point you want. By setting the total force to zero and the torque to zero you will get two equations for the two things you do not know, which are how much weight the ground is supporting and how much weight the man is supporting. That should get you started.

 

[konichiwa2x]

ok here goes

let F1 and F2 be the forces on the bar due to the ground and the man respectively. Let B be the centre of the bar. A and C are the points on the ground and on the man respectively.

so W = F1 + F2
Net torque must be 0.
F1 x AB = F2 x BC
F1.AB.sin(a) = F2.BC.sin(a)
F1=F2 (AB = BC)

so W = 2F1=2F2

F2 = F1 = W/2

Is this correct?

I am sorry.. This must be irritating but I still don't understand how to do this part of the problem :

If the same iron bar were balanced by two men, what will be the be the weight of the bar as experienced by one of the men at the instant the other man let's go of the bar?

thanks for your time.

 

[OlderDan]

ok here goes

let F1 and F2 be the forces on the bar due to the ground and the man respectively. Let B be the centre of the bar. A and C are the points on the ground and on the man respectively.

so W = F1 + F2
Net torque must be 0.
F1 x AB = F2 x BC
F1.AB.sin(a) = F2.BC.sin(a)
F1=F2 (AB = BC)

so W = 2F1=2F2

F2 = F1 = W/2

Is this correct?

I am sorry.. This must be irritating but I still don't understand how to do this part of the problem :

If the same iron bar were balanced by two men, what will be the be the weight of the bar as experienced by one of the men at the instant the other man let's go of the bar?

thanks for your time.

First part looks good. For the second part, whatever force that man is exerting does not contribute to the torque about his point of contact. The torque that is present will cause the bar to rotate, accelerating its center. At the first instant, this is a linear acceleration downwards. Newton's second law applies.