- #1

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

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net torque = 0

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

## The Attempt at a Solution

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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