Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  •  A block A of mass 4 kg is placed on another block B of mass 5 kg, and the block B rests on a smooth horizontal table. If the minimum force that can be applied on A so that both the block

 A block A of mass 4 kg is placed on another block B of mass 5 kg, and the block B rests on a smooth horizontal table. If the minimum force that can be applied on A so that both the blocks move together is 12 N, the maximum force that can be applied on B for the blocks to move together will be :

Option: 1

30 N


Option: 2

25 N


Option: 3

15 N


Option: 4

48 N


Answers (1)

best_answer

When the force of 12 N is applied on block A,

FBD of A is

Equation of motion of A is

12 - f = 4a ...(i)

FBD of B is

Equation of motion of B is

f = 5a ...(ii)

Dividing (i) and (ii), we get

\begin{array}{l} \frac{12-\bar{f}}{f}=\frac{4}{5} \\ \\ f=\frac{20}{3} N=\text { Frictional Force } \end{array}

When force is applied on B:

Frictional force acting between the blocks will remain same. 

FBD of A is:

Equation of motion of A is

f = 4a ...(iii)

FBD of B is

Equation of motion of B is
F - f = 5a ...(iv)

Dividing (iv) by (iii), we get

\begin{array}{l} \frac{\dot{F}-f}{f}=\frac{5}{4} \\ 4 F=9 f \\ F=\frac{9}{4} f=\frac{9}{4} \times \frac{20}{3}=15 N \end{array}

Posted by

shivangi.bhatnagar

View full answer

Similar Questions

Latest Question