A box, mass 25 kg, is placed on a flat, hotizontal surface. There is friction. T

Question

A box, mass 25 kg, is placed on a flat, hotizontal surface. There is friction. The coefficient of kinetic friction between the box and the surface is =0.2. The box ix pulled by a force of Fp=70 N using a cord that makes a 37 degree angle with the horizontal.

a) Compute the weight of the box and the normal force FN between it and the surface. Is this normal force equal to the weight? If so, why? Justify your answer using Newtons 2nd Law in the vertical direction.

b) compute the frictional force Ffr that hte box experiences as it moves right

c) use Newtons 2nd Law to find the acceleration experienced by the box. What forces cause this acceleration?

Explanation / Answer

A. The weight of the box is (25 kg)(9.81 m/s2) = 245.25 N but that is not the normal force. Look at the vertical components: mg, directed down, and Fpsin 37° directed up.

So net vertical force is (25 kg)(9.81 m/s2) - (70 N)(0.601815) = 203.12 N directed down, which is also the normal force. The weight has been reduced by the vertical component of the force that is pulling the box.

B. Horizontal components are Fpcos 37° in the direction of travel, and 0.2 * normal force against the direction of travel. The frictional force is (0.2)(203.12 N) = 40.624 N.

C. Net horizontal force is (70 N)(0.798636) - (0.2)(203.12 N) = 15.28 N in the direction of travel. Acceleration is F/m = 15.28 N / 25 kg = 0.6112 m/s2 to the right.   The horizontal component of the pulling force minus the friction causes this acceleration.

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