Question

A weight weighs 14 lb and rests on the center of the frictionless surface of the ramp supported in equilibrium by ropes AC and BC. Determine the normal force the ramp exerts on the puck as well as the tension in each of the two ropes.

Answer 1

Answer:

Normal reaction = 14lb or

= 14 × 4.44822162 N

= 62.275N (since 1lb = 4.44822162N

T = 7lb or

= 7 × 4.44822162

= 31.138N

Explanation:

The answer above is based on these assumptions

That (1) The ropes are vertically holding the ramps

By Newton's law (third law)

R(Normal reaction) = W

Also, since the ropes are assumed vertical; 2T = W. ----> T = W/2

However if the ropes are at angle a1 and a2, and the let the vertical tensions T1 and T2

T1 = Tsin(a1) also T2 = Tsin(a2)

Hence Tsina1 + Tsina2 = W

T(sina1 + sina2) = W

T = W/(sina1 + sina2)