A chair of mass 12.0 kg is sitting on the horizontal floor; the floor is not frictionless. You push on the chair with a force F = 40.0 N that is directed at an angle of 37.fJ' below the horizontal and the chair slides along the floor. (a) Draw a clearly labeled free-body diagram for the chair. (b) Use your diagram and Newton's laws to calculate the normal force that the floor exerts on the chair.
(a)
(b)
We can break the applied force into its components parallel and perpendicular to the floor:
$F_{\parallel} = F\sin37^\circ = 23.8$ N
$F_{\perp} = F\cos37^\circ = 30.6$ N
The only forces acting on the chair are the force of gravity, $F_g = mg$, and the normal force, $F_n$, from the floor. Since the chair is not accelerating vertically, the normal force must be equal in magnitude and opposite in direction to the force of gravity:
$F_n = F_g = mg = 12.0\text{ kg}\cdot9.81\text{ m/s}^2 = 117.7\text{ N}$
In the horizontal direction, the applied force is the only force, so we can use Newton's second law:
$F_{\parallel} = ma$
$23.8\text{ N} = 12\text{ kg}\cdot a$
$a = 1.98\text{ m/s}^2$
Therefore, the normal force exerted by the floor on the chair is 117.7 N.