Question

Jake drags a sled of mass 18 kg across the snow. The sled is attached to a rope that he pulls with a force of 70 N at an angle of 40°. If there is no friction, what is the normal force acting on the sled?

Answer 1

If the 30 N on the rope were pulled straight up, it would offset the force of gravity ( m g = 10 kg * 9.8 N/kg = 98 N) , leaving a net force up from the ground on the sled of 98-30 = 68 N. Since the rope is pulled at the angle of 25o, only part of the force is in the upward direction, (30N)(sin(25) = (30)(.423) = 12.7. So the net force becomes 98 N down offset by 12.7 up or 98-12.7 = 85.3 N. Ah, there it is: C.

Answer 2

Answer: The normal force acting on the sled 131.4048 Newtons.

Explanation:

On breaking up force of 70 N into vertical and horizontal components:

Vertical components:$f\cos\theta$

Horizontal components:$f\sin\theta$

The normal force is force acting perpendicular on an object.It is a balanced force.

$mg=N+f\sin\theta$

$N=mg-f\sin\theta=18 kg\times 9.8 m/s^2-70 N\times 0.6427=131.4048 N$

The normal force acting on the sled 131.4048 Newtons.