Question

As a logging truck rounds a bend in the road, some logs come loose and begin to roll without slipping down the mountainside. The mountain slopes upward at 48.0 ∘ above the horizontal, and we can model the logs as solid uniform cylinders of mass 575 kg and diameter 1.20 m .

What is the magnitude f of the friction force on the rolling logs?

Answer 1

Since Cylinder is rolling down without slipping, So we need to consider both linear and rotational motion of cylinder. Now Using force balance on the solid cylinder along the incline:

F_net = W*sin - Ff

W = Weight of sphere = m*g

Ff = friction force on sphere

From Newton's 2nd law: F_net = m*a

m*a = m*g*sin - Ff

a = g*sin - Ff/m

Now Using torque balance on the sphere:

= I* = Ff*R

I = Moment of inertia of solid cylinder = (1/2)*m*R^2

= angular acceleration = a/R (when sphere rolls down without slipping), So

(m*R^2/2)*(a/R) = Ff*R

a = 2*Ff/m

So, Now equation both function of acceleration:

a = 2Ff/m = (g*sin - Ff/m)

3Ff/m + Ff/m = g*sin

3*Ff/m = g*sin

Ff = m*g*sin /3

Using given values:

Ff = 575*9.81*sin 48 deg/3

Ff = 1397.3 N = 1400 N = friction force on the sphere

Let me know if you've any query.