Need help understanding forces

Question

5. Find the tension in the cable and the reaction at A. The poles each weigh 2 lbs/ft of length.

6. Find the tension in the cable and the reaction at A. Assume the pole weighs 2 lbs/ft of length.

I am currently a sophmore in aerospace engineering and I am just asking for someone to help explain it to me how the forces are reacting and also the tension on the cable if you could just solve it that would be great as well because I can usually figure it out.

here are the pictures for the questions:

http://home.eng.iastate.edu/~dhaugli/aere160/problem-sets/ps5.html

Answer 1

for question 5

Sum of all forces =0

2 unknowns requires two linearly independent equations.

In this case the forces in the y direction and the moment about point A.

Y=0=-124 - sin45*Tension

Moment about A = 6 ft*sin45*Tension - 36 lbft - 100 lbft - 300 lbft

Solve and find that load at is 180 lb vertically, and the tension in the cable is 79 lb

Answer 2

Force problems are solved by summing all the force components, vertical and horizontal.  If there is a pivot point as at point A in these problems, you should also sum all the moments about that pivot and set to zero.

The moment around point A from the cable tension is T*s, where s is the perpendicular distance from A to the cable.  s = 6*sin(45º), so the moment from the cable is T*6*sin(45º).  The other moment acting at A is from the 100 lbf weight and the poles.  The center of mass of the weight is 4 ft horizontally from the pivot, so its moment is 4*100 lbf-ft.  The center of mass of the horizontal pole is at 3' from the pivot, and its weight is 6*2 lbf.  The moment from the pole is then 3*12 lbf-ft

The moments from the pole and weight are clockwise, and the moment from the tension is CCW, so is negative.  They must all sum to zero, so

4*100 + 3*12 - T*6*sin(45º) = 0

solve for T.

To get the reaction forces at A, you must sum the vertical and horizontal force components from each weight  The vertical component from the cable tension is T*sin(45º), the other vertical forces (all downward) are the weight of the vertical pole, the weight of the horizontal pole and the 100 lbf weight.  Add all of these together to get the vertical reaction force at A.  The only horizontal force is from the tension in the cable, and is T*cos(45º) and that is the horizontal reaction force at A.