Calc III Tension Problem

Question

Ropes 3 m and 5 m in length are fastened to a holiday decoration that is suspended over a town square. The decoration has a weight of 10 kg. The ropes, fastened at different heights, make angles of 52° and 40° with the horizontal. Give your answers correct to two decimal places.

Answer 1

Since you are not clear on what exactly you are trying to find and don't specify which angle goes with which rope, I will give the solution technique in general terms, assuming that you are looking for the tension in each rope. Define T₃ = tension in 3 m rope; T₅ = tension in 5 m rope, the respective angles are labeled below, angle A is horizontal angle with T3 and angle B is the angle wrt horizontal for T5. We are assuming the ropes are "massless". Define the direction of T3 and down as the positive directions.

Consider the forces acting in the horizontal and in the vertical. Since the decoration is not moving (at least not due to the tension in the ropes), the net force in each direction is zero. The force of gravity on the decoration acts only in the vertical direction; the tension in the ropes has components in both the horizontal and vertical.

Horizontal Forces:

Fx = 0 = T3 * cos A - T5 * cos B --> T3 = (T5 * cos B) / (cos A) (alternately, you could solve for T5, it won't matter in the end)

Vertical Forces:

Fy = 0 = mg - T3 * sin A - T5 * sin B

==> mg = T3 * sin A - T5 * sin B

substitute the value for T3 from the horizontal force equation:

==> mg = [ (T5 * cos B) * sin A] / (cos A) + T5 * sin B

Multiply T5 * sin B by (cos A) / (cos A) to get a common denominator and factor out T5:

==> mg = T5 * (cos B sin A + cos A sin B) / (cos A)

using the trig identity sin (A + B) = sin A cos B + sin B cos A this becomes:

mg = T5 * [ sin (A + B)] / (cos A)

Solving for T5

T5 = (mg * cos A) / [sin (A + B)]

once you have T5, you can substitute back into the equation for T3 to get that value also. Remember, which angle is A and which is B will affect your answer.