1. A crate, which has a mass of m = 45.0 kg, is being...

1. A crate, which has a mass of m = 45.0 kg, is being accelerated at 3.20 m/s² up a frictionless, inclined plane at an angle of 35° by a rope as shown to the right. a. Complete the free body diagram showing all forces acting on the crate as it moves up the incline. b. What will be the magnitude of the normal force acting on the crate? c. What will be the magnitude of the tension in the rope? 2. A crate, which has a mass of m = 65.0 kg, is sliding down an inclined plane which has a coefficient of kinetic friction of ?k = 0.43 as shown to the right. The angle of incline is 27°. a. Complete the free body diagram showing all forces acting on the crate. b. What is the magnitude of the normal force acting on the crate? c. What is the magnitude of the friction force acting on the crate? d. What is the magnitude of acceleration of the crate as it slides down the ramp?

Transcript

00:01 Hi, here in this given problem there are two questions based upon the motion of crate on inclined plane.

00:07 In the first one, there is an inclined plane making an angle theta with the horizontal.

00:20 And this theta is given as 35 degree.

00:27 The crate over this inclined plane is being pulled up with the help of a rope, tension in the rope, t, mass of the of the crate if it is m.

00:38 So its weight, mg acting vertically down.

00:42 Component of the weight perpendicular to this incline plane, that is mg, cos theta, normal force exerted by the inclined over the crate, that is also equal to mg cos theta.

00:59 And component of this weight along the inclined plane in downward direction, that is m .g sine theta.

01:07 This is the free body diagram of the crate.

01:12 Here the crate is moving up with an acceleration a given as 3 .20 meter per second square.

01:22 Mass of the crate that is 45 .0 kilogram.

01:28 So in the first part of the problem we have to draw free body diagram of the crate and the and that is already has been drawn.

01:44 So this is as shown in the figure alongside.

01:51 In the second part of the problem, we have to find normal force exerted by the inclined plane over the crate.

02:02 And that is mg coseta means this is 45 multiplied by 9 .8 times cosine of 13.

02:14 T 5 degree and that normal force is calculated to be equal to 361 .2 newton.

02:24 Answer for the second part of this given problem now provided the acceleration is given to us we have to find tension t in the rope.

02:34 So using free body diagram of the crate, the net force acting on the crate that is t minus m g sine theta that should be equal to m a using newton's second law of motion.

02:47 So this tension t will be given by m bracket g sine theta plus a.

02:55 So this is 45 kilogram for g 9 .8 sine 35 degree plus for acceleration of the crate 3 .20.

03:07 So this tension comes out to be equal to 396 .9 newton...

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