# Static Equilibrium with a beam and cable

Juniper7

## 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

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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Homework Helper
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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

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.

Juniper7
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?

Homework Helper
Gold Member
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.

Juniper7
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?