Question

You place a mass m = 40 g on a book and slowly increase the angle e that the book makes with the horizontal. At a critical angle theta_c = 42°, the mass starts to slide so you hold the angle constant. The mass takes 0.75 s to slide a distance of 17 cm to the edge of the book. Assume that the values of mu_s and mu_k, are uniform all over the book.

c) What is the value of mu_s?

d) What is the value of mu_k?

Answer 1

c.)

at critical angle(),

By force balance on mass in inclined direction,

m*g*sin() - fr_s = m*a

here, a = 0 (Since mass starts to slide at critical angle)

m = mass 40 g = 0.040 kg

g = 9.81 m/s^2

fr_s = static friction force = s*N (By force balance on mass in perpendicular direction of inclined plane.)

N = normal force = m*g*cos()

= 42 deg (given)

So, fr_s = m*g*sin()

s*N = m*g*sin()

s*m*g*cos() = m*g*sin()

s = tan() = tan(42 deg)

s = 0.900

d.)

Now when box starts accelerating, then

Using Force balance in inclined direction:

F_net = Fg - fr_k = m*a eq(1)

here, Fg = force due to gravity = m*g*sin()

fr_k = k*N

For acceleration(a), using second kinematics law in inclined direction,

S = u*t + 0.5*a*t^2

given, S = distance = 17 cm = 0.17 m

u = initial speed = 0

t = time = 0.75 s

So, a = 2*S/t^2 = 2*0.17/0.75^2

a = 0.604 m/s^2

then, from eq(1),

m*g*sin() - k*(m*g*cos()) = m*a

k*(m*g*cos()) = m*g*sin() - m*a

k = tan() - a/(g*cos()) = tan(42 deg) - 0.604/(9.81*cos(42 deg))

k = 0.818