First slide

Mechanical equilibrium force cases

Question

A mass M kg is suspended by a weightless string. The horizontal force required to hold the mass at 60^o with the vertical is

Moderate

A

Mg
B

Mg √3
C

Mg(√3 + 1)
D

Mg/√3

Solution

By Lami's Theorem,
$\large \frac{F}{{\sin {{120}^0}}} = \frac{T}{{\sin {{90}^0}}} = \frac{{{F_g}}}{{\sin ({{90}^0} + {{60}^0})}}$
$\large \therefore F = \frac{{{F_g}.\sin {{120}^0}}}{{\sin ({{90}^0} + {{60}^0})}} = \frac{{Mg.\sin {{60}^0}}}{{\cos {{60}^0}}} = Mg\tan {60^0} = \sqrt 3 Mg$

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