Please wait...
Contact Us
Need assistance? Contact us on below numbers

For Study plan details

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

For Franchisee Enquiry



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

93219 24448 / 99871 78554

Mon to Sat - 10 AM to 7 PM

A system of masses is shown in fig with masses and coefficients of friction indicated .Calculate the max value of F for which there is no slipping anywhere,the minimum value of F for which B slides on C,the minimum value of F for which A slips on B.

Asked by m.nilu 16th August 2018, 7:43 PM
Answered by Expert
1. no Slip anywhere 
whole system moves with acceleration a, which is given by,  a = (F-μN)/(20+30+40) = (F-88.2) / 90 m/s2 ................(1)
where μ is the friction coefficient between ground and block-A. N is the normal reaction force between the surfaces of ground and block-A
from the free body diagram given for block-A, we get the no slip condition for block-A is ( m× a )< f ,
where f is frictional force between the surfaces of A and B.
m× a < μ×m×g .........(2)
using (1) and (2), we can write (F-88.2) < 90×μ×g  or ( F-88.2) < 90×0.1×9.8  or F < 176.4 N .................(3)
Hence if the applied force F is less than 176.4 N, there will not be any slip between blocks.
2. Block-B sliding on Block-C
Free body diagram of block-B is given in figure.
we can see that if applied force exceeds sum of firctional forces f1 and f2 , block-B slides on block-C
F >(f1 + f2) = (0.1×20 + 0.2×50)×9.8 = 117.6 N
3. block-A slips on Block-B
As we already calculated for 'no slip anywhere', block-A slips if applied force exceeds 176.4 N
Answered by Expert 27th August 2018, 3:47 PM
Rate this answer
  • 1
  • 2
  • 3
  • 4
  • 5
  • 6
  • 7
  • 8
  • 9
  • 10

You have rated this answer /10

Your answer has been posted successfully!

Chat with us on WhatsApp