Question #162196

A body with the mass m 5 kg lies on an inclined plane and is connected to a solid cylinder with a mass M 10 kg by a massless rope (see sketch). The rope is wound on the solid cylinder. The inclined plane has a static friction of 0.12 and a sliding friction of 0.06. The body is at rest at the beginning.

a) At what critical angle can the body overcome static friction?

b) At the critical angle the body of mass m begins to slide down the plane. Its acceleration is slowed down by the rolling of the full cylinder. How big is the acceleration of the body and how big is the tensile force in the rope?


Expert's answer

As the mentioned diagram is not attached in the question, so i am assuming the given situation



Mass of the body which is on the inclined plane (m) = 5kg

Mass of the cylinder (M)=10kg

Static friction (μs)=0.12(\mu_s) =0.12

Kinetic friction (μk)=0.06(\mu_k)=0.06

Let the angle of inclination of the inclined plane be θ\theta

when body was just going to slip then mgsinθ=μsmgcosθmg\sin\theta = \mu_smg\cos\theta

tanθ=μs=0.12\tan\theta = \mu_s=0.12

θ=6.84\theta = 6.84

mgsin6.84T0.06×mgcos6.84=ma(i)mg\sin 6.84-T-0.06\times mg \cos6.84=ma ----(i)

T.R=IαT. R = I\alpha

T=Ma2(ii)T=\frac{Ma}{2} -----(ii)

From equation (i) and (ii)

mgsin6.84Ma20.06×mgcos6.84=mamg\sin 6.84-\frac{Ma}{2}-0.06\times mg \cos6.84=ma

5×10×sin(6.84)0.06×5×10×cos(6.84)=5a+5a\Rightarrow 5\times 10\times \sin (6.84)-0.06\times 5\times 10 \times \cos(6.84)=5a+5a

10a=\Rightarrow 10a = 50(0.6)

a=3.0m/sec2a=3.0 m/sec^2

T=10×32=15NT=\frac{10\times 3}{2}=15N


LATEST TUTORIALS
APPROVED BY CLIENTS