A sled of mass m = 70.0 kg is being pulled horizontally on a frictionless floor by a force F = 420 N that makes an angle 20.0o above the horizontal. What is the magnitude of the normal force applied by the ground on the sled?
M*g = 70 * 9.8 = 686 N. = Wt. of sled.
Fn = Mg-F*sin20 =
To find the magnitude of the normal force applied by the ground on the sled, we need to analyze the forces acting on the sled.
First, let's resolve the applied force F into its horizontal and vertical components. The horizontal component is Fx = F * cosθ, where θ is the angle above the horizontal. In this case, θ = 20.0°. So,
Fx = 420 N * cos(20.0°)
= 420 N * cos(0.3491) (using radians mode)
= 397.1 N (rounded to one decimal place)
The vertical component of the force F is Fy = F * sinθ.
Fy = 420 N * sin(20.0°)
= 420 N * sin(0.3491)
= 144.7 N (rounded to one decimal place)
Since there is no vertical acceleration (the sled is not sinking into the floor or flying off) and the vertical forces must balance out, the normal force provided by the ground is equal to the vertical component of the applied force.
Therefore, the magnitude of the normal force applied by the ground on the sled is 144.7 N.