00:01
Hello, here we have to solve the following problem.
00:05
The surface on which two blocks sits is frictionless and mass of this blocks is m1 on the bottom, which is 5 and m2 of the top, which is 2 kilogram.
00:16
Co -efficient of friction between these two blocks is 0 .47.
00:22
And in question a, horizontal force f is applied to the bottom block.
00:27
And we have to calculate the maximum force f, can be before m2 begins to sleep to sleep from the m1 let's make the sketch so here let's look at the forces which are acting between m1 and m2 so there is friction force acting by block 2 on 1 and the same friction is acting by 1 on 2 so therefore here we can write down the equation of motion of the block m1.
01:17
For m1, f minus m minus friction force equals to m1 times acceleration.
01:28
And for the block m2, m2 times acceleration must be equal to the friction force, m2a.
01:40
So now, from the second equation, let's express acceleration.
01:46
Acceleration then is friction force divided by m2.
01:50
And here this static friction force is coefficient of static friction times m2g over m2.
01:59
So therefore, f minus static friction force, which is mu coefficient of static friction m2g, must be equal to m1 times acceleration, which is coefficient of static friction m1g...