Question

1. a 735 n mountain climber is rappelling down the face of a vertical cliff as shown in the diagram. if the rope makes an angle of 12° with the cliff face, what is the tension in the rope?

Explanation:

Step1: Analyze the force components

The weight of the climber \(F_g = 735N\) acts vertically down - ward. The tension \(T\) in the rope has a vertical component. Since the climber is in equilibrium (not accelerating in the vertical direction), the vertical component of the tension balances the weight of the climber. The vertical component of the tension is \(T\cos(12^{\circ})\) (using the adjacent - side formula for the angle with the cliff face in a right - triangle of force components).

Step2: Set up the equilibrium equation

In the vertical direction, \(\sum F_y=0\). So \(T\cos(12^{\circ})=F_g\).

Step3: Solve for \(T\)

We know that \(F_g = 735N\) and \(\cos(12^{\circ})\approx0.9781\). Then \(T=\frac{F_g}{\cos(12^{\circ})}\). Substituting the values, \(T=\frac{735}{0.9781}\approx752.5N\).

Answer:

The tension in the rope is approximately \(753N\).