Question

Jake drags a sled of mass 25 kg across the snow. The sled is attached to a rope that pulls with a force of 50 N at an angle of 32 degrees. If there is no friction, what is the normal force acting on the sled? ​

Answer 1

Answer:

218.5 N

Explanation:

In order for the sled to be in equilibrium along the vertical direction, the forces acting along this direction must be balanced. So the equilibrium equation is:

$N+F sin \theta - mg =0$

where

N is the normal force

F = 50 N is the force that pulls the sled

$\theta$ is the angle between the force and the horizontal, so

$F sin \theta$ is the component of F acting along the vertical direction

(mg) is the weight of the sled, with

m = 25 kg being the mass of the sled

g = 9.8 m/s^2 is the acceleration due to gravity

Solving the formula for N, we find

$N=mg-F sin \theta = (25 kg)(9.8 m/s^2)-(50 N)sin 32^{\circ}=218.5 N$