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?

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.

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:

Horizontal components:

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

The normal force acting on the sled 131.4048 Newtons.

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