Cable tension statics problem

[Shay10825]

1.) A uniform rod AB is 1.2 m long and weighs 16 N. It is suspended by vertical strings AC and BD. A block P weighing 96 N is attached at E, .3 m from A. The tension in the string BD is?
http://img116.exs.cx/img116/2822/STATICS1.jpg
I did:

(.3m)(96N)+ (.9m)(16N) - (1.2 m)(x) = 0
x=36N

But the answer is 32 N

What did I do wrong??

2.) A 960 N block is suspended as shown. The beam is weightless and is hinged to the watt at A. The tensions in the cable BC is?
http://img116.exs.cx/img116/364/statics2.jpg

I did:

(4m)(960N)=t
t=3840

The answer is 1600 N.

~Thanks

 

[Diane_]

In your first answer, you got the arithmetic wrong - I used the same pivot point you did and got 32N.

In your second, you need to do a little trigonometry. The vertical component of the tension is 960N. Figure out what the total tension needs to be to get a vertical component of that magnitude. (Hint: you can get the angle at B with an Arctangent function.)

 

[wais]

In the first problem, you have correctly set up the equation for the sum of forces in the vertical direction. However, you have made a mistake in calculating the moment arm for the weight of the rod (16N). The moment arm should be 0.6m, not 0.9m. This gives you a final answer of 32N, which is the correct answer.

In the second problem, you have correctly set up the equation for the sum of moments about point A. However, you have not taken into account the weight of the beam itself. The weight of the beam creates a clockwise moment about point A, which needs to be subtracted from the total moment in order to find the tension in the cable. The correct equation should be:

(4m)(960N) - (2m)(x) - (2m)(160N) = 0

Solving for x gives you a tension of 1600N, which is the correct answer.