Contact
Need assistance? Contact us on below numbers

For Enquiry

10:00 AM to 7:00 PM IST all days.

Business Enquiry (West & East)

OR

or

Thanks, You will receive a call shortly.
Customer Support

You are very important to us

For any content/service related issues please contact on this number

8788563422

Mon to Sat - 10 AM to 7 PM

# A 12 N of force required to applied on A to slip on B.find maximum horizontal force F so that both block both A and B move together

Asked by ashutosharnold1998 26th November 2019, 1:01 AM

Left most figure shows the free body diagram of block-A when 12 N force is applied to block-A
and block-A is about to slip on block-B.

hence we have , μN = ( μ 4 g ) = 12  or  μ = 12/ (4 × 9.8 ) = 0.306

where μ is friction coefficient, N is normal force equals 4g , where g is acceleration due to gravity.

Middle figure shows the forces acting on block-A , when block-B is pulled with force Fm .

Now friction force μN pulls block-A. Hence acceleration of block a = (μN) / 4 = (0.306×4 ×9.8)/4 = 3 m/s2

if both the blocks move together, acceleartion of system is a = 3 m/s2

Right most figure shows the free body diagram of block-B , when system moves with a = 3 m/s2

friction force shown as μN is the reaction force on block-B due to friction acting on block-A

Hence, if Newton's law is applied to block-B, we have

Fm - μN = mB × a   or Fm = mB × a + μN = ( 5 × 3 + 0.306 × 4 × 9.8  ) = 27N
Answered by Expert 26th November 2019, 7:33 AM
• 1
• 2
• 3
• 4
• 5
• 6
• 7
• 8
• 9
• 10

You have rated this answer /10

Tags: force