The Force supported by the Pin at point O

Northbysouth

Homework Statement

To test the deflection of the uniform 270-lb beam the 120-lb boy exerts a pull of 35 lb on the rope rigged as shown. Compute the force supported by the pin at the hinge O.

I have attached an image of the question.

ƩMO = 0
ƩFx =0
ƩFy =0

The Attempt at a Solution

My first thought is to find the moment about point O, for which I could then solve for the force at point A.

0 = -A(1.9') + (270lb)(5.45') + T(8.5') + (120lb)(7.0')

A = 1373.157 lb

However I'm not sure whether I have the right center of gravity of the beam, because I'm unsure whether the ball at point A (or whatever that is) alters it. So would the center of gravity of the beam be at 5.45' or at 4.5'?

Also, am I correct in assuming that the 35lb force that the boy applies to the rope is transferred completely to the rope that is 8.5' away from O?

Assuming that I am correct, I did the following then found the force in the y direction at the point O, Fo

0 = Foy +Ay -270lb -120lb -35lb
Foy = -Ay + 270 + 120 + 35
F0y = -948.157895

I tried entering 948.15 as I had thought that perhaps it wanted the magnitude of the force but it didn't accept this.

Attachments

• Boy on a beam.png
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Gold Member
My first thought is to find the moment about point O, for which I could then solve for the force at point A.

0 = -A(1.9') + (270lb)(5.45') + T(8.5') + (120lb)(7.0')

A = 1373.157 lb

Consider the 120 pound boy just standing on the beam and another person, who is on a platfrom not on the beam, pulling on the rope. How does that change your forces in the y-direction, or does it? Have you taken all the forces in the y-direction with the boy pulling on the rope?

pongo38
Consider a cut and replace the cut with the force or forces revealed. Can you see a cut that would enable your moment equation to yield a valuable result? I think you have very nearly done this, but there is one force missing from your equation. It really does help to actually draw the free body diagram about which you are making equilibrium statements.

Northbysouth
256bits: Are you saying that the 35lb force that the boy exerts is operating in both the positive y direction and negative y direction? I'm fairly sure that there is a 35lb force acting in the negative y direction at 8.5' from O, but is there also a 35lb force at 10.9' acting in the positive y-direction?

pongo38: I'm don't understand what you mean by cuts, but the only force that I can think of that could be missing would be a 35lb force acting at 10.9' from point O in the positive y direction.

pongo38
The whole diagram given is a 'cut' whose boundaries are the reactions (at 4 places in this case). If you cut horizontally where the 2.4' dimension is, you have to replace the cut with two rope forces pulling down, and you are still left with the reactions at A and at O. So, yes, the force missing is the 35 lb force DOWNWARDS 10.9' from O. The slanting 35 lb force is not cut and therefore does not feature in the moment equation. It might help to draw a free body diagram of the boy to understand all the forces that keep him in equilibrium. Can you then see what equilibrates the horizontal component of the slanting rope force?

Northbysouth
pongo38: you were right. I reversed the direction of that force at the 10.9'. Altering my calculations appropriately I got 1113.9 lb as my answer which is correct.

ekwan
Sorry in regards to the mass of the beam and person, did you guys take the gravity into account to get the weight? A little confused about that. I am currently doing a similar problem and I have been assuming that the weight of both the person and the beam had to be taken into account.

pongo38
ekwan, I think you would be correct to take account of the weight of the beam, because it has been given, and it is significant in relation to the weight of the boy in this case.