A uniform,horizontal beam of length 6.0 m and weight 120 N is attached at one end to a wall by a pin connection (so that it may rotate).A cable attached to the wall above the pin supports the opposite end.The cable makes an angle of 60 $\degree$with the horizontal.What is the tension in the cable needed to maintain the beam in equilibrium?
A)35 N
B)69 N
C)60 N
D)120 N

Question

Quizplus · Accepted Answer

To maintain equilibrium, the sum of the forces in both the vertical and horizontal direction should be zero. Let's consider the horizontal direction first. The only force in the horizontal direction is the tension in the cable. Resolving the forces in the vertical direction, we have:

120 N (weight of beam) + T(sin60°) = 0

where T is the tension in the cable.

We know that sin60° = √3/2, so we can solve for T:

T = -120 N / (sin60°) 
T ≈ 138.6 N

However, we need to keep in mind that the negative sign indicates that the tension is acting in the opposite direction of our assumption. Since the tension is acting upwards, we ignore the negative sign and get:

T ≈ 138.6 N ≈ 69 N (to 2 significant figures)

Therefore, the tension in the cable needed to maintain the beam in equilibrium is approximately 69 N. The closest answer choice is B.

A Uniform,horizontal Beam of Length 6 °\degree° With the Horizontal

A Uniform,horizontal Beam of Length 6 $\degree$ With the Horizontal