How do I find that the tensions from both sides are equal?

[link223]

Homework Statement

A man having a weight of 875 N attemps to hold himself using one of the two methods shown. Determine the total force the must exert on bar AB in each case and the normal reaction he exerts on the platform at C . The platform has a weight of 150 N.

Relevant Equations

Newton's laws

• Q5 Medium level: How do I know that both tension forces from both sides are the same? Visually I might conclude that because of the symmetry but how do I see that mathematically? (/with the equilibrium eqs. to be more specific)

Thanks in advance

 

[Lnewqban]

The problem is asking about the total force on the bar AB, regardless of how centered or not his hands and feet are respect to it.

 

[link223]

The problem is asking about the total force on the bar AB, regardless of how centered or not his hands and feet are respect to it.

Can you help me a little bit out on how to start this problem then please? Because at every way I look at it and cut the figure (for FBD's) I always keep getting those T_a and T_b back so I need some relation for those I would presume?

 

[kuruman]

How about showing your work? Maybe you missed something.

 

[link223]

How about showing your work? Maybe you missed something.

I will, thanks. just charging my phone and I'll drop the picture.

 

[link223]

@kuruman

 

[haruspex]

How do I know that both tension forces from both sides are the same?

If you assume the arrangement is symmetric, you can deduce that equality by taking moments. But as @Lnewqban implies, you do not need to assume symmetry. The two tensions should always appear in the linear force equations (as opposed to torque equations) as their sum. How that sum is split between the two does not matter here.

 

[link223]

If you assume the arrangement is symmetric, you can deduce that equality by taking moments. But as @Lnewqban implies, you do not need to assume symmetry. The two tensions should always appear in the linear force equations (as opposed to torque equations) as their sum. How that sum is split between the two does not matter here.

oohh okay, thank you, I see! but.. Why can we make the assumption that I could just sum them up (talking about the tensions from side A and side B)?

 

[kuruman]

Your FBDs for the pulleys and the equations that you get from them look OK. I am not so sure about the rest because you have more FBDs than equations and that is a sign of trouble. Another sign of (possible) trouble is the down arrow labeled W_M. Is that the weight of the man? If so, the weight of the man is not the force exerted by the man on the platform. It is the force exerted by the Earth on the man. I hope you see the difference.

If you wish to receive additonal help, please draw 3 separate FBDs: (1) for the bar that the man holds, (2) the man and (3) the platform on which he stands. Underneath each FBD write the equation that is obtained from it legibly. Also label each force in a way that it will be understood by third parties. This will make it easier for said parties to help you. Thanks.

 

[haruspex]

oohh okay, thank you, I see! but.. Why can we make the assumption that I could just sum them up (talking about the tensions from side A and side B)?

It is not an assumption; it is just how things will turn out in the equations. You can see that must happen by considering the force balance on the person+platform system.

 

[link223]

Your FBDs for the pulleys and the equations that you get from them look OK. I am not so sure about the rest because you have more FBDs than equations and that is a sign of trouble. Another sign of (possible) trouble is the down arrow labeled W_M. Is that the weight of the man? If so, the weight of the man is not the force exerted by the man on the platform. It is the force exerted by the Earth on the man. I hope you see the difference.

If you wish to receive additonal help, please draw 3 separate FBDs: (1) for the bar that the man holds, (2) the man and (3) the platform on which he stands. Underneath each FBD write the equation that is obtained from it legibly. Also label each force in a way that it will be understood by third parties. This will make it easier for said parties to help you. Thanks.

Thank you. I totally confused the force on that one my bad.. it was getting a bit late lol...
But I drew all of those FBD's to find some kind of relation for the tension in the ropes once again. Otherwise, I only had 3 but without a relation for tensions I would've been stuck.(which haruspex explained)

 

[link223]

It is not an assumption; it is just how things will turn out in the equations. You can see that must happen by considering the force balance on the person+platform system.

Could you explain this one please? I don't really see how I would be able to see that one from the person + platform C's force equation as there are other forces working as well.

But I do think however that It could be obtained from the moment equation of platform C itself, because if I look at the drawing (the actual one from the question because mine is wrong) we have a normal force from the man on the platform and the weight of the platform itself. and that on the two ends we have the tension which will equal each other from the moment equation obviously.

 

[haruspex]

I don't really see how I would be able to see that one from the person + platform C's force equation as there are other forces working as well.

Let the combined weight be W and the tensions be $T_A, T_B$. For (b), what equation do you get?

 

[Lnewqban]

Can you help me a little bit out on how to start this problem then please? Because at every way I look at it and cut the figure (for FBD's) I always keep getting those T_a and T_b back so I need some relation for those I would presume?

The problem considers two pulley systems only to ensure stability, but you could imagine one single central pulley system for simplification of your analysis, and then, you could do your FBD and balance equations using only centrally located forces and reactions.