# Equilibrium problem - Vertical beam, cable

• Valerie Prowse
In summary: Determine the reaction forces acting on the beam by the groundThe reaction forces acting on the beam are Rx = 520N, Ry = -520N, and τ = 0 N.
Valerie Prowse

## Homework Statement

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

∑Fx = 0
∑Fy = 0
∑τ = 0

## 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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Valerie Prowse said:

## Homework Statement

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

∑Fx = 0
∑Fy = 0
∑τ = 0

## 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.
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.

PhanthomJay said:
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.

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?

Valerie Prowse said:
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?
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.

Valerie Prowse said:

## Homework Statement

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

∑Fx = 0
∑Fy = 0
∑τ = 0

## 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!

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.

Satvik Pandey said:
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##.
View attachment 80882

I hope this helps.

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?

Valerie Prowse said:
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?
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.

PhanthomJay said:
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.

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?

Exc
Valerie Prowse said:
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?
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.

PhanthomJay said:
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.

Great! Thank you everyone for your help!

## 1. What is an equilibrium problem in relation to a vertical beam and cable?

An equilibrium problem in relation to a vertical beam and cable refers to the state in which the vertical beam and cable are in balance, with no external forces causing them to move or change position.

## 2. How is equilibrium achieved in a vertical beam and cable system?

Equilibrium in a vertical beam and cable system is achieved when the sum of all the forces acting on the system is equal to zero. This means that the forces pulling the beam and cable in different directions are balanced, resulting in a stable and stationary system.

## 3. What factors can affect the equilibrium of a vertical beam and cable?

Several factors can affect the equilibrium of a vertical beam and cable system, including the weight and length of the beam and cable, the angle at which the cable is attached to the beam, and any external forces such as wind or additional weight on the beam.

## 4. How does the tension in the cable change when the angle of the cable is altered?

The tension in the cable changes when the angle of the cable is altered because the angle affects the direction and magnitude of the force acting on the cable. As the angle increases, the tension in the cable also increases, and vice versa.

## 5. What are some real-world applications of equilibrium problems in vertical beam and cable systems?

Equilibrium problems in vertical beam and cable systems are commonly found in construction and engineering, such as in the design of bridges, cranes, and other structures. They are also relevant in physics and mechanics, as they demonstrate the principles of balanced forces and stability.

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