# Equilibrium problem - Vertical beam, cable

1. Mar 23, 2015

### Valerie Prowse

1. The problem statement, all variables and given/known data
A uniform vertical beam of mass 40 kg is acted on by a horizontal force of 520 N at its top and is held, in the vertical position, by a cable as shown.

a. Draw a free-body diagram for the beam, clearly labeling all of the forces acting on it.
b. Calculate the tension in the cable
c. Determine the reaction forces acting on the beam by the ground

2. Relevant equations
∑Fx = 0
∑Fy = 0
∑τ = 0

3. The attempt at a solution
What I am confused about (reading other people's posts trying to work through this question) is I think I have a force missing somewhere. Some people are talking about friction force or some kind of hinge force at the bottom of the beam... I'm not sure where they are getting that information from, though. I've been reviewing problems in my textbook that are similar (they're all horizontal beams hanging from cables and hinged on a wall, though) and I can see how there could be a hinge/reaction/something force coming from the ground, but I really don't completely understand the logic behind it.

∑Fx = 0
0 = FT⋅cos28 - F
(If this were true, then you could just solve for FT, but I don't think that is correct)
If there is something missing, I feel like it would be some sort of hinge force pointing some way towards the cable, and it could be broken down into components as well... so maybe:
∑Fx = 0 = FT⋅cos28 + FH⋅cosθ - F

and

∑Fy = 0
0 = FN + FH⋅sinθ - mg - FT⋅sin28

... I'm not really entirely sure about torque yet. I'm just looking for some guidance on where to get started on this question! Thanks!

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2. Mar 23, 2015

### PhanthomJay

there has to be a horizontal ground reaction force on the beam or see it would tip over. This could be a friction force or a hinge force if the beam is pinned to the ground. And there has to be a vertical reaction force from ground on beam, or else it would sink into the ground as if it were a hole. Did you draw the free body diagram of the beam? What forces, both known and unknown, act on it, including the ground forces? Then apply your relevant equilibrium equations, including the torque equation.

3. Mar 23, 2015

### Valerie Prowse

Does this mean that the reaction force is different from the normal force?

I did draw the FBD, but I am unsure in which direction the reaction force would be pointing. I thought maybe, from looking at different types of questions, that it would be pointing in the direction of the cable? Or directly upwards?

4. Mar 23, 2015

### PhanthomJay

Don't worry about in which direction the ground reaction force points. It will have an x and y component Rx and Ry. Assume a direction for Rx left or right, and Ry up or down. Continue with your equations.

5. Mar 24, 2015

### Satvik Pandey

As the center of mass of the beam is at rest so Net force in vertical direction is zero. Now as two downward forces are acting on the beam (mg and a component of tension) so it is fair to assume that an upward component of Normal reaction acts on it.

Now to find tension you can equate net torque acting on the beam about the point (at which the beam is in contact with the ground) to zero. This will give you a equation from which $T$ can be found easily.
You will find that force acting on the beam ($F=520N$) is greater than the horizontal component of tension acting the beam. But the Center of mass of the beam is in equilibrium. So a force should act in horizontal direction(rightward) so that net force in horizontal direction becomes zero. Equate them to get an equation.

I forgot to mark $mg$.

I hope this helps.

6. Mar 25, 2015

### Valerie Prowse

Ahh this makes sense.

So, the torque equation about the bottom of the beam would be:
Fp = 520N

∑τ = 0 = Fp ⋅ 5m - FT⋅cos28⋅3m - FN(x)⋅0m
0 = Fp ⋅ 5m - FT⋅cos28⋅3m
FT = Fp⋅5 / cos28⋅3
FT = 2600/2.64 = 984.8N

However, this is much more than Fp... and the horizontal component of FT would also be much more than Fp? Have I forgotten a component when calculating torque?

7. Mar 25, 2015

### PhanthomJay

no, your equation is very good. (The torque of the beam weight is also 0). So yes the tension force is correct. Now solve for the x and y forces at the base of the beam. Make sure you determine the correct direction for both.

8. Mar 25, 2015

### Valerie Prowse

I worked out the rest of the question in case this was okay:

∑Fy = FNy - mg - FT⋅sin28
FNy = FT⋅sin28 + mg = (984.8⋅sin28) + (40⋅9.8) = 462.3 + 392 = 854.3N

∑Fx = FT⋅cos28 + FNx - Fp = 0
Fp - FT⋅cos28 = FNx
FNx = 520 - (984.8⋅cos28) = -349.5
(This makes me think that FNx is pointing left, instead?)

FN = √(FNy2 + FNx2)
FN = √(-349.5)2 + (854.3)2 = 923.0N

θ = tan-1(y/x) = tan-1(854.3/-349.5) = -67.8°
Or rather, 67.8° from the horizontal, but in the leftwards direction?

9. Mar 25, 2015

### PhanthomJay

Exc
Excellent, all correct. Note that leaving the base reactions as x and y components rather than a resultant force at an angle is acceptable and often preferred.

10. Mar 25, 2015

### Valerie Prowse

Great! Thank you everyone for your help!!