Question

A box of books weighing 319 N is shoved across the floor by a force of 485 N exerted downward at an angle of 35 degres below the horizontal.a) If the coeficent of friction between the box and the floor is 0.57, how long does it take to move the box 4 meters, starting from rest?b) If If the coeficent of friction between the box and the floor is 0.75, how long does it take to move the box 4 meters, starting from rest?

Answer 1

Let w, n, p, and f denote the magnitudes of the 4 forces acting on the box.

• w = weight = 319 N

• n = normal force

• p = pushing force = 485 N

• f = friction = µ n, where µ is the coefficient of static friction

The net force on the box points in the direction that the box moves, which is to the right. In particular, this means the box is vertically in equilibrium. Split up the vectors into their vertical and horizontal components, and apply Newton's second law. (I take up and right to be the positive vertical and horizontal directions, respectively.)

• vertical:

p sin(-35°) + n - w = 0

and solving for n,

- (485 N) sin(35°) + n - 319 N = 0

n ≈ 597 N

• horizontal:

p cos(-35°) - f = m a

where a is the magnitude of the net acceleration on the box. Solve for a. Since f = µ n and m = w / g (where g = 9.80 m/s² is the mag. of the acc. due to gravity) we get

p cos(35°) - µ n = (w / g) a

(485 N) cos(35°) - µ (597 N) = (319 N) / (9.80 m/s²) a

a ≈ (12.2 - 18.3 µ) m/s²

(a) If µ = 0.57, then the net acceleration on the box is

a ≈ (12.2 - 18.3 • 0.57) m/s² ≈ 1.75 m/s²

so that the time t required to move the box 4 m is

4 m = 1/2 a t ²

t ≈ √((8 m) / (1.75 m/s²))

t ≈ 2.14 s

(b) The box does not move.

If µ = 0.75, then

a = (12.2 - 18.3 • 0.75) m/s² ≈ -1.55 m/s²

but a negative acc. here means the applied acc. points *opposite* the direction of movement, thus making the box move backward which doesn't make sense. The coefficient of friction is too large for the given applied force to get the box moving. With µ = 0.75, the frictional force to overcome has mag. f ≈ 448 N. But the given push contributes a horizontal force of (485 N) cos(-35°) ≈ 397 N. This mag. needs to be increased in order to get the box moving.

Answer 2

(a) The time taken to move the box 4 meters is 2.14 s.

(b) The box will decelerate when the coefficient of friction is 0.75 and cannot be moved to 4 meters forward.

The given parameters;

• weight of the book, W = 319 N
• applied force, F = 485 N
• angle of inclination, θ = 35 ⁰

The mass of the books is calculated as;

$m = \frac{W}{g} \\\\m = \frac{319}{9.8} \\\\m = 32.55 \ kg$

The normal force on the box is calculated as follows;

$F_n = -W - Fsin\theta\\\\F_n = -319 - (485\times sin35)\\\\F_n = -597.18 \ N$

The frictional force when the coefficient of friction is 0.57;

$F_f = \mu F_n\\\\F_f = 0.57 \times -597.18\\\\F_f = -340.39 \ N$

The acceleration of the box is calculated as follows;

$F cos \theta - F_f = ma\\\\(485)cos(35) \ - 340.39 = 32.55 a\\\\56.899 = 32.55a\\\\a = \frac{56.899}{32.55} \\\\a = 1.75 \ m/s^2$

The time taken to move the box 4 meters is calculated as;

$s = v_0t + \frac{1}{2} at^2\\\\s = 0 + \frac{1}{2} at^2\\\\t = \sqrt{\frac{2s}{a} } \\\\t = \sqrt{\frac{2\times 4}{1.75} } \\\\t = 2.14 \ s$

(b) The frictional force when the coefficient of friction is 0.75;

$F_f = \mu F_n\\\\F_f = 0.75 \times -597.18\\\\F_f = -447.885 \ N$

The acceleration of the box is calculated as follows;

$F cos \theta - F_f = ma\\\\(485)cos(35) \ -447.885 = 32.55 a\\\\-50.596 = 32.55a\\\\a = \frac{-50.596}{32.55} \\\\a = -1.55\ m/s^2$

Thus, the box will decelerate when the coefficient of friction is 0.75 and cannot be moved to 4 meters forward.

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