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Homework Statement
A load Q is applied to the pulley C, which can roll on the cable ACB. The pulley is held in position shown by a second cable CAD, which passes over the pulley A and supports a load P. Knowing that P=750N, determine
a) the tension in cable ACB
and
b) the magnitude of load Q
Homework Equations
Using cos and sin rules...?
The Attempt at a Solution
I drew a free body diagram around C, with Q going down, a cable going to the right 25 degrees above the horizontal and a cable going to the left 55 degrees above the horizontal.
The cable on the left i assumed to be 750N.
So the sum of the forces along the X axis is (Tension of ACB)cos25-750cos55=0 and T came out to equal 474.7N.
The sum of the forces along the Y axis is (Tension of ACB)sin25+750sin55-Q=0 and I got Q to equal 241.6N
The answers came out to be in the kiloNewtons.
I also tried another way after finding out my answers were completely off. If it is 750N down, then the cable going from C to A would equal 915.6N (CA = 750/sin55). Using 915.6 as the force for my left cable on my free body diagram, the sum of the forces along the X axis is (Tension of ACB)cos25-915.6cos55=0 and the Tension came out to be 579.5N.
The sum of the forces along the Y axis is 579.5sin25+915.6sin55-Q=0 and Q=994.9N. These answers were also wrong.
So please, can anyone direct me in the correct direction?
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