Torque of board over 2 shoulders

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Homework Help Overview

The problem involves two men carrying a 2 m long board with a mass of 71 kg, while an object exerting a force of 200 N hangs from one end. Participants are tasked with determining the forces acting on each man.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the need for a free body diagram and the concept of rotational physics. There is uncertainty about how to distribute the total force among the two men and how to apply torques in the context of the problem.

Discussion Status

Some participants have sketched diagrams and attempted to identify the center of mass of the board. There is a recognition of the role of torques and fulcrums in the problem, but confusion remains regarding the calculations and the application of relevant formulas. Guidance has been offered regarding the use of torque equations and the relationship between forces and distances.

Contextual Notes

Participants express difficulty with the textbook's explanations and the lack of clarity in the problem setup. There is mention of specific distances related to the forces acting on the board, but the exact formulas needed for calculations are not provided in the discussion.

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Homework Statement


Two men are carrying a board that is 2 m long and 71 kg mass. Man one is at the end of board mand 2 is 0.6 m from other end and hanging from the end is and object which has a force of 200 N. What are the forces on the two men?


Homework Equations



Also says to draw a free body diagram?

The Attempt at a Solution



The instructor has not really gone over free body diagrams so am not sure where to go with that portion.

I understand how to figure out the force when the equation is balanced but do not know where to begin when there are two points.

I figured out that the Total Force would be 71 kg x 10m/s2 = 710 N + 200 N = 910 N but am not sure how to distribute it among the two men.
 
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Welcome to Physics Forums.

A free body diagram is just a sketch that shows the forces as arrows from the points they act upon. You've probably seen these before (maybe under the name "force diagram").

Do you see that there is rotational physics going on here?
 
Yes that is the section of the book I am in is rotational motion and understand the rest of thed questions and we have the answers in the back of the book but am not sure where they are going with this one problem
 
verden said:
Yes that is the section of the book I am in is rotational motion and understand the rest of thed questions and we have the answers in the back of the book but am not sure where they are going with this one problem

Sketch it. Be sure to include the weight of the board acting at the board's center of mass. Then treat each man as a fulcrum when trying to figure out the force applied to the other man.

See how far you can get with that.
 
I have the drawing of the two men and the center of mass of the board between the two men is 710 N so I see that the one man is more of a fulcrum than the other as he has the mass of .6m of the board and 200 N hanging off the end and .4 m to the center of the boards mass in the book it says that he has 793 N and that leaves 117 N on the other guy but I don't see the math. Is there only really the one guy acting as a fulcrum because the weight on the end of the board is actually pulling down on the board and bringing the other end up off the other man
 
verden said:
I have the drawing of the two men and the center of mass of the board between the two men is 710 N so I see that the one man is more of a fulcrum than the other as he has the mass of .6m of the board and 200 N hanging off the end and .4 m to the center of the boards mass in the book it says that he has 793 N and that leaves 117 N on the other guy but I don't see the math. Is there only really the one guy acting as a fulcrum because the weight on the end of the board is actually pulling down on the board and bringing the other end up off the other man

It's not that one is more of a fulcrum than the other. Man 1 is a fulcrum when calculating the forces on Man 2. Man 2 is a fulcrum when calculating the forces on Man 1. That 710 N is what acts on Man 2. Here is why:

Man 2 is pushing up with some force on the plank to stop it from rotating about the shoulder of Man 1. He pushes up at r = 1.4 m from Man 1, providing a torque that counters the two opposing torques (from the 710 N force at r = 1 and the 200 N force at r=2). If you write that equation out and solve for the force applied upward by Man 2, you will get 793 N.

Since (by the third law of motion) the plank pushes on man one with the same force he pushes on the plank, he is pushed down with 793 N.

Go through the math and see that it works. Then think of the center of rotation being at Man 2 and find what force Man 1 must apply up to counter the net torque.
 
I see where you are coming from with the torques and the force in the rotations but I do not understand where the r = 1 and r = 2 come into play and also the book does not give me the formulas I need to understand how to work the problem. These books have been hard to follow but am able to get most of it through the problem solving book on the other questions but this one it is not giving me a clue can you pleaese let me know the formula to use. Thanks
 
verden said:
I see where you are coming from with the torques and the force in the rotations but I do not understand where the r = 1 and r = 2 come into play and also the book does not give me the formulas I need to understand how to work the problem. These books have been hard to follow but am able to get most of it through the problem solving book on the other questions but this one it is not giving me a clue can you pleaese let me know the formula to use. Thanks

Just as a net force causes a translational acceleration, net torque causes a rotational acceleration. The formula is
[tex] \bold{\tau}\ =\ \bold{r}\times\bold{F}[/tex]
Where r is the distance from the center of rotation that the force is applied, and F is perpendicular to r.

When we treat Man 1 as the center of rotation the problem you posted, THere are two forces acting on the board that will causes torques:
[tex]\tau[/tex]1=710N x 1 m = 710 N-m
[tex]\tau[/tex]2= 200n x 2 m= 400 N-m

Man 2 has to apply a force where he is to create a torque equal and opposite to the sum of those two torque (i.e., 1110 N-m). Write the expression for that torque and solve for force.

(It is about midnight where I am. I'll be signing off for about 6 hours. Good luck.)
 

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