A 150-newton force, applied to a wooden crate at an angle of 30° above the horizontal, causesthe crate to travel at constant velocity across a horizontal wooden floor.

1. Calculate the magnitude of the normal force exerted by the floor on the crate. [Show all work, including the equation and substitution with units.]2. Determine the magnitude of the frictional force acting on the crate. 3. Calculate the magnitude of the horizontal component of the 150-newton force. [Show all work, including the equation and substitution with units.]

A component of 150 N in vertical direction will reduce the magnitude of reaction force.

reaction force exerted by the floor

= mg - 150 sin 30

where m is mass of the crate.

the magnitude of the horizontal component of the 150-newton force

150 cos30

= 130 N

This force tries to pull the crate in forward direction with acceleration but it has no acceleration. It is so because frictional force of equal magnitude acts on it in opposite direction which makes the net force acting on it equal to zero.

Hence frictional force is equal to 150 cos 30.

= 130 N.