First slide

Mechanical equilibrium force cases

Question

Consider three blocks A,B and C. Block A is placed on block B, which is placed on block C and block C is placed on the ground. Normal reaction between blocks B and C is three times of that between blocks A and B. Normal reaction between block C and ground is two times of that of between blocks B and C. Possible values of the masses of blocks A , B and C respectively are :

Easy

A

5 kg, 15 kg and 10 kg
B

5 kg, 10 kG and 15 kg
C

5 kg, 15 kg and 30 kg
D

5 kg, 30 kg and 60 kg

Solution

$\begin{array}{l} N_{AB} = m_{A} g \\ N_{BC} = (m_{A} + m_{B}) g = 3 m_{A} g \Rightarrow m_{B} = 2 m_{A} \\ N_{CG} = (m_{A} + m_{B} + m_{C}) g = 6 m_{A} g \Rightarrow m_{C} = 3 m_{A} \\ \Rightarrow If m_{A} = 5 kg, then m_{B} = 10 kg and m_{C} = 15 kg \end{array}$

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