Find the tension in the cable and reactive forces at 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

 

[TSny]

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]

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?

 

[TSny]

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

 

[jfnn]

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

Thank you!