# Find the tension in the cable and reactive forces at ground

• jfnn
In summary, the conversation discussed a problem involving a utility pole with a horizontal force acting at its top and held in place by a cable. The problem required finding the tension in the cable and the reaction forces exerted by the ground at the lower end of the pole. The solution involved setting up a free-body diagram and using equations for net torque and net force to find the necessary values. The conversation also addressed the unknown direction of the friction force and how to approach finding the horizontal reaction force from the ground.
jfnn

## Homework Statement

A 160-kg utility pole extends 12m above the ground. A horizontal force of 250 N acts at its top and the pole is held in the vertical position by a cable, as shown in the figure (I have attached the photo).

1) draw a free-body diagram for the pole

2) Determine the tension in the cable

3) What are the reaction forces exerted at the lower end of the pole by the ground?

## Homework Equations

[/B]
net torque = 0

net force = 0 (in x and y direction as well)

## The Attempt at a Solution

[/B]
a) The free body diagram:

1. Weight acts downwards at the center of the pole
2. Normal force acts upwards from the bottom of the pole
3. Tension acts in the same direction of the rope (shown in diagram) pulling outwards
4. The horizontal force applied (as shown in the diagram) acts in the same direction at the top
5. The friction (however, I am not sure what direction the friction acts because I do not know the direction of motion if the pole tips?) Any suggestions?

b) The tension is found by:

I defined the axis of rotation (o) at the bottom of the pole --> Therefore, the torque by the normal force, friction, and weight is zero

Thus, net torque = torque of horizontal force (Tf) - torque of tension (Tt) (with clockwise being negative)

Therefore,

Tf = r * F*sin (theta) --> Tf = 12 * 250 * sin (90) --> 3000 N*m
Tt = r* F * sin(theta) --> Tt = 8 * T * sin 55

net torque must equal zero for equilibrium --> Therefore,

0 = 3000 N*m - (6.5532T)
-3000 = -6.5532T
T=458 N --> Is this math right/approach correct?

c) The reactive forces at lower end of pole by ground is

1. The force of friction (would it be kinetic friction?)
I have no clue how to calculate this force...
I know friction kinetic = uk * N
I found N below, therefore --> kinetic friction = uK * (1.83*10^3)
How do I find uK?

2. Normal force
N = W + Ty (tension in y direction) --> N = mg + Tsin(theta) --> N = (160)(9.8)+458sin(35) --> N = 1.83 *10^3 N

#### Attachments

• Screen Shot 2017-08-06 at 8.33.29 PM.png
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The work that you've done so far looks good to me.

The friction force is an unknown. So, you don't know its direction at the beginning. You can just assume a direction and set up your equations according to this assumption. If calculation yields a negative value for the force, then you know it is actually in the opposite direction of your initial choice.

The friction would not be kinetic since the pole is presumably not sliding across the ground. In fact, the force is not necessary friction. The pole could be secured to the ground in some way. You can just call the force the "horizontal component of the reaction force from the ground". There is no need to worry about coefficients of friction.

To find the horizontal component of the reaction force, use the same method as you did to find N.

jfnn
TSny said:
The work that you've done so far looks good to me.

The friction force is an unknown. So, you don't know its direction at the beginning. You can just assume a direction and set up your equations according to this assumption. If calculation yields a negative value for the force, then you know it is actually in the opposite direction of your initial choice.

The friction would not be kinetic since the pole is presumably not sliding across the ground. In fact, the force is not necessary friction. The pole could be secured to the ground in some way. You can just call the force the "horizontal component of the reaction force from the ground". There is no need to worry about coefficients of friction.

To find the horizontal component of the reaction force, use the same method as you did to find N.

Thank you for your response. So the two forces are the normal force, which I found. Then the second would be the horizontal reactive force from the ground.

Since I know the sum of the forces in the x direction must equal zero, then I can say (with the east direction positive):

The horizontal reactive force of ground (f) + the tension in the x direction (Tx) - the horizontal applied force (F) = 0

f + Tx - F = 0 where f is unknown, F = 250 N, and Tx = 458*sin(55)

Thus,

f + 458*sin(55) - 250 = 0
f = -125.17

So... since it is negative, it is in the opposite direction that I originally defined... Therefore, it is pointing to the left at the bottom of the pole?

Looks good. Yes, the horizontal reaction force on the pole is to the left.

jfnn
TSny said:
Looks good. Yes, the horizontal reaction force on the pole is to the left.
Thank you!

## 1. What is tension in a cable?

Tension is the force that is transmitted through a cable or rope when it is pulled tight. It is the force that keeps the cable in a state of equilibrium and maintains its shape.

## 2. How is tension in a cable calculated?

Tension in a cable can be calculated using the formula T = F * L, where T is the tension, F is the force applied to the cable, and L is the length of the cable.

## 3. What factors affect the tension in a cable?

The tension in a cable is affected by the magnitude of the force applied, the material and thickness of the cable, the angle at which the cable is pulled, and the weight of any objects attached to the cable.

## 4. What is the difference between tension and compression in a cable?

Tension refers to the force that is pulling the cable, while compression refers to the force that is pushing the cable. Tension is typically caused by external forces, whereas compression is caused by the weight of the cable itself or any objects attached to it.

## 5. How do reactive forces at ground affect the tension in a cable?

Reactive forces at ground can affect the tension in a cable by providing an opposing force to the applied force, resulting in a balance of forces and a stable tension in the cable. The magnitude and direction of the reactive forces will depend on the angle and direction of the cable's tension.

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