The three blocks m1 29 kg m2 4 kg and m3 77 kg lie on a...

Transcript

00:01 So in this problem here we have three masses lying on a frictionless surface and we have a force affecting on mass one here and we need to find what is the force exerted on block three by block two? so first if you consider this first diagram we can say that our f here our force our force acts all of this it moves the three masses so they act all as one body so it's m1 plus m2 plus m3 times their acceleration according to newton's second law and if we rearrange this equation we can get the acceleration as the force f over m1 plus m2 plus m3 and we're going to keep this for later use okay so here now for the second diagram here which is the free by diagram of forces on ms one we have the force on mass 1 due to mass 2 and our initial force here those 14 newtons so now we can say we can apply newton's second as well so if we assume the direction of motion is in this direction in the right direction so it would be f minus f1 2 is equal to mass 1 times acceleration so if we can rearrange this for f1 2 so if we can rearrange this for f1 2 so if 1 2 would be f minus m1a sorry m1a however we have a here from the previous step so we can now say that f12 is just f minus m1 times f over the submission of the masses and we can simplify this further to be if you multiply the course multiply here so we're just left with f f1 .2 is equal to m2 plus m3 times f divided by m1 plus m2 plus m3 which is the summation of masses here.

02:26 Good.

02:28 However, we also know that f12 is the same as f2 1.

02:34 Let's just keep this in mind here.

02:36 Now we have one diagram remaining which is the third one here.

02:40 We can do the same analysis.

02:42 On it so if we assume that action of motion is to the right so now we have f21 minus f2 3 it's just equal to m2 times a and now we rearrange this equation for m2 3 sorry f2 3 which is the force on m2 exerted by on due to m3 this would be f2 1 minus m2 times a and we have f2 1 from the previous step here so you can substitute this in and we also know that this whole part is a so i'm just going to keep it a here for simplicity so it's going to be m2 plus m3 times a okay so we can substitute this that means that f two three is just m2 plus m3 times a minus m2 times a minus m2 times so we can see that these can sell out together...

Question

The three blocks, m1 = 2.9 kg, m2 = 4 kg and m3 = 7.7 kg lie on a frictionless surface. Determine the magnitude of the force exerted on block 3 by block 2 if F = 14 N.

Question

The three blocks, m1 = 2.9 kg, m2 = 4 kg and m3 = 7.7 kg lie on a frictionless surface. Determine the magnitude of the force exerted on block 3 by block 2 if F = 14 N.

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