Exercise 2.31 cylinder and hanging mass

A uniform cylinder of mass MM sits on a fixed plane inclined at an angle θ\theta. A string is tied to the cylinder’s rightmost point, and a mass mm hangs from the string, as shown in Figure above. Assume that the coefficient of friction between the cylinder and the plane is sufficiently large to prevent slipping.

What is mm, in terms of MM and θ\theta, if the setup is static?

#Mechanics #Statics

Note by Beakal Tiliksew
7 years ago

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In the above image, I have marked out the forces on the cylinder, and also, I have removed the mass m\displaystyle m.

Firstly, if the system is static, then the tension in the string must be mg\displaystyle mg, so that there is equilibrium for the mass m\displaystyle m.

For attaining rotational equilibrium,
Considering the torque about the contact point of the cylinder and the inclined plane,

Mg(Rsinθ)=T(RRsinθ)Mg(R\sin\theta) = T(R-R\sin\theta)

The torque due to the Normal Reaction and the Frictional force, is zero, since those forces pass through the contact point (the point which we are considering the torque about)

Mg(Rsinθ)=mgR(1sinθ)\Rightarrow Mg(R\sin\theta) = mgR(1-\sin\theta)

Msinθ=m(1sinθ)\Rightarrow M\sin\theta = m(1-\sin\theta)

m=Msinθ1sinθ\Rightarrow \boxed{m = \frac{M\sin\theta}{1-\sin\theta}}

Anish Puthuraya - 7 years ago

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Anish, which software do you use to draw diagrams?

Karthik Kannan - 7 years ago

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I use Photoshop CS5

Anish Puthuraya - 7 years ago

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@Anish Puthuraya Oh, I don't have it. Could you recommend something else with which I can put a decent diagram?

Karthik Kannan - 7 years ago

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@Karthik Kannan Paint is not bad, if you just draw simple diagrams.

Anish Puthuraya - 7 years ago

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@Anish Puthuraya Thanks, Anish! : )

Karthik Kannan - 7 years ago

nice explanation

Hafizh Ahsan Permana - 6 years, 11 months ago
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