Question

A book of mass 5 kg rests on a plank. You tilt one end of the plank and slowly increase the angle of the tilt. The coefficient of static friction between the book and the plank is 0.50. What is the maximum angle of tilt for which the book will remain stationary and not slide down the plank?

Answer 1

Answer:

The maximum angle of tilt is 26.56°.

Explanation:

Given that,

Mass of book = 5 kg

Coefficient of static friction = 0.50

The normal force of plank on book

$F_{n}=mg\cos\theta$

The friction force holding book against moving

$f_{\mu}=\mu F_{n}$....(I)

The force induced by gravity is

$F_{g}=mg\sin\theta$....(II)

We need to calculate the angle

Now, both forces are equal

$f_{\mu}=F_{g}$

$\mu\ mg\cos\theta=mg\sin\theta$

Put the value into the formula

$0.50\times5\times9.8\cos\theta=5\times9.8\sin\theta$

$\tan\theta=\dfrac{0.50\times5\times9.8}{5\times9.8}$

$\theta=\tan^{-1}{0.5}$

$\theta=26.56^{\circ}$

Hence, The maximum angle of tilt is 26.56°.

Answer 2

Answer: The maximum angle of tilt is 26.57°

Explanation:

mass of book = 5 kg

weight = mass × acceleration due to gravity

wgt of book = 5 × 9.8 = 49 N

coefficient of static friction = Fs = 0.50

let maximum angle of tilt = A

force parallel to the plank = Fp

force perpendicular to the plank = Fn

Fp = 49 × sin A

Fn = 49 × cos A

Fs = u × Fn = 0.5 × 49 × cos A

Fp - Fs = m × a

49 × sin A - 0.5 × 49 × cos A = m × 0 = 0

49 × sin A = 0.5 × 49 × cos A

(divide both sides by 49)

sin A = 0.5 × cos A

(divide both sides by cos A)

sin A/cos A = 0.5

NOTE: sin A/cos A = tan A

replace sin A/cos A with tan A

∴ tan A = 0.5

A =\tan^{-1}{0.5}

A = 26.565°

(note that to get A, take the arctan of 0.5, where arctan is tan raised to the power of -1)

The maximum angle of tilt is 26.57°.