Question

A force of 49N is just sufficient to pull a block of wood weighing 10kg on a rough horizontal surface. Calculate the coefficient of friction and angle of friction.

Answer 1

mass of the wood m = 10 kg

weight of the wood = mg = 10* 9.8 = 98 N

now the normal force vertically upwards on the block will counterbalance the weight so it is W = N = 98 N

now for just move the block it requires 49 N force

So F = 49 N  is static friction on the block

As per the formula of static friction force

$F_f = \mu_s N[tex]</p><p>[tex]49 = \mu_s 98$

$\mu_s = 0.5$

now for the angle of friction we have

$\theta = tan^{-1}\frac{F_f}{N}$

$\theta = tan^{-1}\frac{\mu_s N}{N}$

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

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

$\theta = 26.6 degree$

so the angle of friction is 26.6 degree

Answer 2

Answer:

mass of the wood m = 10 kg

weight of the wood = mg = 10* 9.8 = 98 N

now the normal force vertically upwards on the block will counterbalance the weight so it is W = N = 98 N

now for just move the block it requires 49 N force

So F = 49 N is static friction on the block

As per the formula of static friction force

F_f = \mu_s N[tex] [tex]49 = \mu_s 98F

f

=μ

s

N[tex][tex]49=μ

s

98

\mu_s = 0.5μ

s

=0.5

now for the angle of friction we have

\theta = tan^{-1}\frac{F_f}{N}θ=tan

−1

N

F

f

\theta = tan^{-1}\frac{\mu_s N}{N}θ=tan

−1

N

μ

s

N

\theta = tan^{-1}\mu_sθ=tan

−1

μ

s

\theta = tan^{-1}0.5θ=tan

−1

0.5

\theta = 26.6 degreeθ=26.6degree