Static Equilibrium with a beam and cable

  • Thread starter Juniper7
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  • #1
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Homework Statement


A uniform vertical beam of mass 40kg is acted on by a horizontal force of 520N at its top and is held, in the vertical position, by a cable as shown in the attached picture.
a) Calculate the tension in the cable
b) Determine the reaction forces acting on the beam by the ground

Homework Equations


F=T-mg

The Attempt at a Solution


I think I over simplified this:

∑Fx = Fcablecos28° - Fapplied force = 0
Fccos28° = Fa
Fc = 588.94N <---- answer to a)

∑Fy = Fcsin28° + mg - FN = 0
Fcsin28° + mg = FN
(588.94N)sin28° + (40kg)(9.81m/s2) = FN
FN = 668.49N <----- answer to b)

Thank you in advance for any help!
 

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Answers and Replies

  • #2
PhanthomJay
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Homework Statement


A uniform vertical beam of mass 40kg is acted on by a horizontal force of 520N at its top and is held, in the vertical position, by a cable as shown in the attached picture.
a) Calculate the tension in the cable
b) Determine the reaction forces acting on the beam by the ground

Homework Equations


F=T-mg

The Attempt at a Solution


I think I over simplified this:

∑Fx = Fcablecos28° - Fapplied force = 0
Fccos28° = Fa
Fc = 588.94N <---- answer to a)

∑Fy = Fcsin28° + mg - FN = 0
Fcsin28° + mg = FN
(588.94N)sin28° + (40kg)(9.81m/s2) = FN
FN = 668.49N <----- answer to b)

Thank you in advance for any help!
Although not stated, the beam is presumably supported at its base by a pinned joint which can take both horizontal and vertical forces. You neglected to include the horizontal force at that beam base support. Try summing moments about it to calculate Ty, then continue.
 
  • #3
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Although not stated, the beam is presumably supported at its base by a pinned joint which can take both horizontal and vertical forces. You neglected to include the horizontal force at that beam base support. Try summing moments about it to calculate Ty, then continue.

Ok, So that would be like the force of friction? working in the opposite direction of the applied force?
 
  • #4
PhanthomJay
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Ok, So that would be like the force of friction? working in the opposite direction of the applied force?
We'll it could be friction or a bolt connection , but it's direction wouldn't be opposite the applied force. Try summing moments about a point and summing forces = 0 and see what happens.
 
  • #5
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We'll it could be friction or a bolt connection , but it's direction wouldn't be opposite the applied force. Try summing moments about a point and summing forces = 0 and see what happens.

I'm sorry, I haven't learned to sum moments yet. Or what moments are. Is there a way to do it without that?
 
  • #6
PhanthomJay
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I'm sorry, I haven't learned to sum moments yet. Or what moments are. Is there a way to do it without that?
Oh terminology. Moment is another name for torque. Sum torques like you did last problem. Vert comp of tension force at guy base times it's perp distance to beam base is equal to applied force times it's perp disr to beam base. That's one way to do it....others using free body diagrams....
 

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