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Torque and Forces Question (With a Diagram!)

  • Thread starter Siann122
  • Start date
  • #1
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Homework Statement


https://www.physicsforums.com/attachment.php?attachmentid=40303&stc=1&d=1319521828

An iron ball of mass 220kg is suspended from the end of a rigid steel girder. The girder has a mass of 180kg and a length of 2.4m. The girder is pivoted to a rigid upright. A steel wire is attached 1.6m along the girder. It holds it in equilibrium with angles between component as shown in the diagram.

a). Identify all the forces acting on the girder by drawing them on the diagram.
b). Demonstrate by calculation that the tension in the steel wire is 5390N
c). If the tension in the steel wire is 5390N, calculate the magnitude and direction of the reaction force from the pivot acting on the steel girder.

Homework Equations


F = ma
T = rF
Ʃ = 0

The Attempt at a Solution


I have absolutely no idea how to even attempt this question, I just don't know where to start.
 

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Answers and Replies

  • #2
110
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Draw a free body diagram of the rod. Label all the forces acting on it and break them into x and y components.

Sum the forces in the x and y directions and know they must be equal to zero.

Next sum the torques.
 
  • #3
37
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Draw a free body diagram of the rod.
The diagram drawing itself is confusing me. Does the ball and mg go in the same place (the center of gravity on the rod) or does a seperate arrow need to be drawn in for the ball?
 
  • #4
307
3
Draw a free body diagram of the rod. Label all the forces acting on it and break them into x and y components.

Sum the forces in the x and y directions and know they must be equal to zero.

Next sum the torques.
Great solution! :)
 
  • #5
110
0
The diagram drawing itself is confusing me. Does the ball and mg go in the same place (the center of gravity on the rod) or does a seperate arrow need to be drawn in for the ball?
Draw a free body diagram of the rod only. The ball is not part of this. What is part of this is the tension in the rope where the ball is hanging, that is equal to the balls weight. So you have the weight of the ball acting where the rope is and then you have mg of the rod acting at the rods center of gravity which is half the length of the rod (we have to assume the density is constant if it doesn't say that otherwise it becomes a little more complicated).
 
  • #6
110
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Great solution! :)
lol was that sarcastic? Should I have added anything else?
 

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