CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

In the figure shown, the coefficient of static friction between C and ground is 0.5, coefficient of static friction between A and B is 0.25, coefficient of static friction between B and C is zero. The minimum value of force F (in Newton), to cause sliding between A and B is  (Masses of A,B and C are respectively 2 kg,4 kg and 5 kg).


Solution

Since there is no friction between B and C, there is no possibility of a force acting on C and it will not move.

A and B will move together until A reaches its maximum acceleration and friction between A and B reaches limiting value.
So,
fABL=mA amax
μmAg=0.25(20)=2amax
amax=2.5 ms2
This acceleration is brought about by F in (A + B) system.
F=(mA+mB)amax
F=(2+4)2.5
F=15 N

So, a minimum force of 15 N has to be applied for the block A to slip over block B.

flag
 Suggest corrections
thumbs-up
 
0 Upvotes


Similar questions
QuestionImage
QuestionImage
View More...


People also searched for
QuestionImage
QuestionImage
View More...



footer-image