Exercise 2.31 cylinder and hanging mass

A uniform cylinder of mass \(M\) sits on a fixed plane inclined at an angle \(\theta\). A string is tied to the cylinder’s rightmost point, and a mass \(m\) 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 \(m\), in terms of \(M\) and \(\theta\), if the setup is static?

Note by Beakal Tiliksew
3 years, 10 months ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

alt text

alt text

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

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

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

\[Mg(R\sin\theta) = T(R-R\sin\theta)\]

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)

\[\Rightarrow Mg(R\sin\theta) = mgR(1-\sin\theta)\]

\[\Rightarrow M\sin\theta = m(1-\sin\theta)\]

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

Anish Puthuraya - 3 years, 10 months ago

Log in to reply

nice explanation

Hafizh Ahsan Permana - 3 years, 9 months ago

Log in to reply

Anish, which software do you use to draw diagrams?

Karthik Kannan - 3 years, 10 months ago

Log in to reply

I use Photoshop CS5

Anish Puthuraya - 3 years, 10 months ago

Log in to reply

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

Karthik Kannan - 3 years, 10 months ago

Log in to reply

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

Anish Puthuraya - 3 years, 10 months ago

Log in to reply

@Anish Puthuraya Thanks, Anish! : )

Karthik Kannan - 3 years, 10 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...