A 80 kg climber is standing horizontally on a perfectly vertical cliff face. The climber is 1.8 m tall and is attached by a 2 m long rope fastened around their middle to a point on the cliff above them. What is the normal force the cliff face exerts on the climber in Newtons?
Details and assumptions
The climber is not moving.
The acceleration of gravity is −9.8 m/s2.
The center of mass of the climber is at the point where the rope meets their body.
395.06
396.40
To find the normal force exerted by the cliff face on the climber, we need to consider the forces acting on the climber in the vertical direction.
First, let's draw a free-body diagram to visualize the forces acting on the climber:
1. The weight of the climber acts downward, with a magnitude of (mass) × (acceleration due to gravity). In this case, the weight is given by: weight = mass × acceleration due to gravity = 80 kg × (-9.8 m/s^2) = -784 N (negative sign indicates downward direction).
2. The tension in the rope connects the climber to the cliff face. The rope exerts a force upward on the climber, equal in magnitude and opposite in direction to the weight. Therefore, the tension in the rope is also -784 N.
3. The normal force exerted by the cliff face on the climber acts perpendicular to the cliff face. Since the climber is not moving, the normal force must balance the vertical forces acting on the climber. In this case, the normal force will have the same magnitude as the weight, but in the opposite direction to balance the forces.
Therefore, the normal force exerted by the cliff face on the climber is 784 N in the upward direction.