Question

A box of books weighing 325 N is shoved across the floor by a force of 475 N exerted downward at an angle of 35° below the horizontal. (a) If ????k between the box and the floor is 0. 55, how long does it take to move the box 5. 80 m starting from rest? s (b) What is the maximum coefficient of friction between the box and the floor that allows the box to move from this applied force.

Answer 1

The net forces on the box acting perpendicular and parallel to the floor are

∑ F[perp] = F[normal] - 325 N + (475 N) sin(-35°) = 0

∑ F[perp] = (475 N) cos(-35°) - F[friction] = ma

where m is the mass of the box and a is its acceleration.

Solve for F[normal] :

F[normal] = 325 N + (475 N) sin(35°) ≈ 597 N

Then the frictional force has magnitude

F[friction] = 0.55 F[normal] ≈ 329 N

and so

60.5 N ≈ (325 N) a/g

(note that sin(-35°) = -sin(35°), cos(-35°) = cos(35°), and mg = 325 N so m = (325 N)/g)

Solve for a :

a = (60.5 N) / (325 N) g ≈ 1.82 m/s²

(a) Assuming this acceleration is constant, starting from rest, the box achieves a final velocity v such that

v² = 2a∆x

v² = 2 (1.82 m/s²) (5.80 m)

⇒ v ≈ 4.60 m/s

which happens in time t such that

v = at

4.60 m/s = (1.82 m/s²) t

⇒ t ≈ 0.177 s

(b) Let µ be the coefficient of static friction. The box just begins to slide if the magnitude of the parallel component of the applied force matches the magnitude of friction, i.e.

∑ F[para] = (475 N) cos(-35°) - F[friction] = 0

We have

F[friction] = µ F[normal] = (597 N) µ

so that

(597 N) µ = (475 N) cos(35°)

⇒ µ ≈ 0.651