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Hi, here in this given problem, put on a horizontal frictionless platform, there is a system of three blocks.
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The first block a, the second block b, and the third block c.
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These are the three blocks a, b and c, a force being applied on the block a towards left.
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The magnitude of this force is 8 .25 newton.
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Masses of the blocks are m .a is equal to 6 .90 kilogramm b is equal to 2 .85 kilogram.
01:01
And mc is equal to 1 .50 kilogram.
01:07
So first of all, we find net acceleration in this system towards left, for which we use newton's second law of motion.
01:28
And it says net force acting on a system is equal to the product of mass of the system into its acceleration.
01:39
So here this acceleration will be given by.
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A is equal to for the force this is 8 .25 and 4 the addition of these masses here it is 11 .25.
01:57
So this acceleration here comes out to be 0 .73 meter per second square.
02:07
Now we have to find the contact forces between the blocks.
02:11
So, first of all, block a will exert a force fba on the block b in return as a reaction as per newton's third law of motion, block b will exert a force f, a b on block a.
02:35
Similarly here, block b will exert a force at c, f, c b, and block c will exert a force of force f, b, c, force at b due to c.
02:49
We have to find the values of these forces in the first part of the problem...