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Old Apr 5th 2015, 12:49 AM   #1
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Question Rigid object in equilibrium problem (I think)!

I don't really know how to approach this problem. But I think that it is a rigid object in equilibrium problem. This is it: The step ladder of negligible weight is constructed as shown in the figure. A painter of mass 70.0 kg stands on the ladder 3.0m from the bottom. Assuming the floor is frictionless, find (a) the tension in the horizontal bar connecting the two halves of the ladder. The correct answer is 133 N. I have attached the figure that was included with the problem as well as the work that I did. Somewhere I got mixed up. Thank you.
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File Type: pdf Scan0011.pdf (924.4 KB, 21 views)

Last edited by Newtphys; Apr 5th 2015 at 12:54 AM.
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Old Apr 6th 2015, 10:05 AM   #2
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I find it easiest to first calculate the reaction forces at the left and right feet of the ladder - call these forces R_ A and R_B. You can do this by considering moments acting on the ladder acting about point A:

Sum of Moments at A = 0 = R_B x 2 m - W x 3 m x sin(theta)

This gives R_A, then R_B comes from sum of forces in y-direction = 0.

Now consider the right hand piece of the ladder - it has R_B acting upward, T (tension in the connecting rod) acting to the left, and a horizontal and vertical reaction force at the pivot at the top. Finding the vertical component of the force at the pivot is now trivial, and you can see that the horizontal component must be equal to T, though acting in the opposite direction. Now you can perform a sum of moments about B = 0 and determine a value for T.
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Old Apr 7th 2015, 04:15 AM   #3
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I can't make sense of your answer. I have another attachment here. Is this what you were trying to say?
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Old Apr 7th 2015, 04:42 AM   #4
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Yes - you have determined the reaction force R_B, and can next determine the reaction force at the other leg, R_A.

The attached may explain the steps I suggested a bit clearer.
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Rigid object in equilibrium problem (I think)!-ladder.jpg  

Last edited by ChipB; Apr 7th 2015 at 01:37 PM.
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Old Apr 10th 2015, 05:57 PM   #5
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I came back to this question and still wasn't able to solve it. I think the teacher will post solutions pretty soon, so I will check my work against that of the teacher. The answer to the question is 133 Newtons. I attached the work I did to go a bit further with this problem. The teacher provides 133 Newtons as the answer and it is up to the student to show the procedure used to get the answer.
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Old Apr 11th 2015, 05:32 AM   #6
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At the top of your 3rd page you use sin(phi) in your calculation of torque about point A. Should be cos(phi).
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