00:03
Here we have a system of three blocks which is being pushed by a force f and we have to find out the magnitude of contact force that m1 exerts on m2.
00:14
Now for each of the box there will be a friction force so let us find the friction force first so friction force is opposite to motion small f and normal forces fn and gravitational force is m g downward so in vertical direction there is no acceleration so normal force must balance the weight and so friction force will be equal to mu coefficient of friction times normal force equal to mu m g so from this argument the friction force on m1 is equal to f1 mu m g similarly friction force on m2 is equal to mf1 m m2g and friction force on f3 is equal to mu m3g.
01:20
Now when these blocks are being pushed by a force of 80 newton, they move together so they will have a common acceleration.
01:31
Let us call this acceleration a.
01:33
Since they have common acceleration, we can take these three blocks as a system.
01:39
So the mass of the system is let us say capital m and the forward force is f capital f and the friction force is f1 plus f2 plus f3 due to the friction force on the three blocks.
02:00
So in horizontal direction we have f1 minus f2 minus f3 equal to mass times acceleration.
02:10
Now we put the values minus mu m1g minus mu m2g minus mu m3g equal to capital m is the mass of all the three blocks so it will be m1 plus m2 plus m3 times a so here we have m sorry f minus mu g can be taken common and we have m1 plus m2 plus m m3 equal to m1 plus m2 plus m3 times a.
02:51
So acceleration a will be equal to f upon m1 plus m2 plus m3 minus mu g...