Question

A rope hangs between two cliffs, edge A and edge B. There is a 250 kg hiker sliding across and is closer to

Answer 1

Answer:

1655.5 N

Explanation:

We resolve the forces on the rope into horizontal and vertical components. Since the only vertical force exerted on the rope is the weight of the man, this weight equals the vertical components of the tensions on both sides of the rope .

The only horizontal forces acting on the rope are horizontal components of both tensions on both sides. Since the hiker doesn't move sideways, they cancel out and he is balanced. So, the net horizontal force is zero.

Let T be the tension in the rope. So, each vertical component on side A is Tsin35° and that on side B is Tsin65°. Let the weight of the man be W = mg where m = mass of hiker = 250 kg and g = acceleration due to gravity = 9.8 m/s².

So, since the hiker balances, Tsin35° + Tsin65° = mg

0.5736T + 0.9063T = 250 kg × 9.8 m/s²

1.4799T = 2450 N

dividing both sides by 1.4799, we have

T = 2450 N/1.4799

T = 1655.52 N

T ≅ 1655.5 N

So, the tension in the rope equals T = 1655.5 N