Statics problem (moment and equilibrium) (1 Viewer)

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1. Having really hard time understanding the concept of moment and equilibrium in statics. In the following picture the man is carrying a 6kg uniform rigid board. I'm asked to calculate the amount of force the man feels on his shoulder. (at point A)

Moment.png


Now I know I'm supposed to start by making a free body diagram, but I don't know where would the weight of the board be placed. Is it in center of the board regardless of the support point at A, or do I have to separate the weight into 2 components based on its length, one in front of the man and one behind him?
 

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berkeman

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Welcome to the PF. :smile:

The worker is having to pull down at point B in order to balance the board on their shoulder. Where is the center of mass of the board? How do the force down at B and the force up at A keep the board from falling or turning?

Please take a try at the FBD so we can work from there. Thanks.
 
Welcome to the PF. :smile:

The worker is having to pull down at point B in order to balance the board on their shoulder. Where is the center of mass of the board? How do the force down at B and the force up at A keep the board from falling or turning?

Please take a try at the FBD so we can work from there. Thanks.
I think I figured it out but I'm not sure.
Moment.png

So ∑M=(W1×(1.5/2))-(W2×(0.9/2)-(0.6B)=0
Then by calculating Force at point B and adding it to the total weight, the total force on the worker shoulder is calculated?
 

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berkeman

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The total free weight should be acting down at the center of mass -- I'm not able to tell if that's what you are doing yet from your FBD...
 
The total free weight should be acting down at the center of mass -- I'm not able to tell if that's what you are doing yet from your FBD...
I seperated the weight:into 2 components. But I also used your method and got the same answer! So I calculated the moment using point A as reference based on your method:
W=6×9.81=58.86 and the weight force is acting at 1.2 meter from the edge or 0.3 meter from worker's shoulder (1.2-0.9=0.3)
Then ΣM=Rd so ΣM= (58.86×0.3) - (0.6B) = 0 ⇒ B= 17.658/0.6 = 29.43
Force on worker's shoulder is then W+B = 58.86 + 29.43 = 88.29
 

berkeman

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Looks reasonable. Be sure to include units in all of your calculations. What are the units of Force or Weight? :smile:
 

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