Two masses m1 50 kg and m2 30 kg are connected by a cord as...

Two masses m1 = 50 kg and m2 = 30 kg are connected by a cord as shown in Figure. The coefficient of kinetic friction between mass m2 and the incline (   = 37°) surface is   k = 0.30. The other mass slides on a frictionless horizontal surface. (a) On the graph shown, draw the free-body diagram for each mass. (6 pts) (b) Calculate the friction force acting on mass m2. (4 pts) (c) Calculate the acceleration of the system of the two masses. (8 pts)

Transcript

00:01 According to newton second law, the sum of forces, which we also call as net force, is equal to the product of the mass and acceleration.

00:11 So here consider a two -block system where one block is on the incline.

00:19 It's called that block as m2.

00:23 And it is connected to a via cord to another mass.

00:32 We call as m1 that is on a horizontal plane.

00:35 So the angle of the incline where m2 is is equal to theta and that's equal to that's put in put here even already.

00:46 Theta is equal to 37 degrees.

00:49 The masses m1 and m2 are 50 and 30 kilograms respectively.

00:58 It is also given that the horizontal surface where m1 is rest is is on is is frictionless.

01:07 Whereas the surface where m2 is with friction that has a coefficient of friction equals to point 30 so what we want to do is to find the free body diagram for letter a or rather to draw the free body diagram in another b we're going to compute for the friction force on m2 or letter c we're going to complete the acceleration of the system so starting with the free body diagram we show the forces that are acting on the object so we'll do it separately for mass 1 and mass 2 so starting with mass 2 so the forces acting on it are no its own weight points downward the normal force that is perpendicular to the plane the friction force we also have the tension between the cord between along the cord between m1 and m2 and we can also so right at here we can also express the weight into components perpendicular and parallel to the plane the one parallel is mg or rather sign data with data that is the same as the data of the inclined and the perpendicular is m2g cosine cosine so this is for the mass two on mass one it's fbd is simpler because it does not have friction force on it and also it's not inclined we what we do have are the normal force the so just to distinguish normal force two here and normal force one the weight which is m1g and the tension force so this is for the free body diagram moving on to letter b we draw the how we solve for the friction force so we know that friction is equal to the product of the coefficient of friction in this case kinetic friction because the object is moving multiplied by the normal force so you see the idea that in the direction perpendicular to the plane we have the forces normal force and opposite to it is the component of the weight perpendicular to the plane which is m2g cosine so that two these two forces have jacketed from it by each other because they are in opposite direction and since m2 is not moving in that direction, the sum of these forces must be 0.

04:01 Therefore, we have n to be equal to m2g cosine theta...

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