First slide

Static friction

Question

A uniform rope of length / lies on a table. If the coefficient of friction is $μ$ , then the maximum length x of the part of this rope which can overhang from the edge of the table without sliding down is

Moderate

A

$l / μ$
B

$l / (μ + 1)$
C

$μl / (μ + 1)$
D

$μl / (μ - 1)$

Solution

Let l be the length of the chain and m be its mass.
$∴$ mass per unit length = (m / l)
Weight of the hanging part $= (m / l) \times x \times g$
This weight provides tension on the portion of the chair lying on the table.
weight of the chain on the table $= (m / l) (l - x) g$
This gives the normal reaction on the table.
Hence the friction p .R is given by
$μR = μ (m / l) (l - x) g$
For maximum length of the hanging part, we have
Tension = Friction
$(m / l) (x \times g) = μ (m / l) (l - x) g$
Solving, we get $x = (\frac{μl}{μ + 1}) .$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App