2.) A box is being dragged up hill as shown in the diagram...

2.) A box is being dragged up hill as shown in the diagram below. There is a pulling force on the box that makes an angle of 30º with respect to the surface of the hill. The mass of the box is 130kg. The angle of the hill is 20º. There is no friction. (40pts) a.) Find the magnitude of pulling force if the box is moving at constant velocity. b.) Find the normal force the hill exerts on the box when it is moving at constant velocity. c.) Find the magnitude of the pulling force when the box has an acceleration of 1.0 m/s^2. d.) Find the magnitude of the normal force the hill exerts on the box when moving at the acceleration in part c.

Transcript

00:01 We are given that there is no friction present.

00:02 This is 20 degree inclination.

00:06 This block has a mass of, i think mass of the block is 1 30 kilogram.

00:14 And we apply a force f that is at an angle of 30 degree with this inclination.

00:21 So in the first part we need to calculate that what is this force f if it moves with constant acceleration, constant velocity.

00:31 If you try to draw the free body diagram, vertical.

00:33 Downward would be 130 .00 this would be normal.

00:39 All right if you break this force into f force 30 and this is f sine 30 right so now it has to go along the inclined plane with constant velocity that means net force along the inclined plane has to be zero so if we take two components of this force this would be 1300 post 20 this would be 1300 sine 20 so, this 1 ,300 sine 20 must be equal to f post 30.

01:13 So f would be equal to 1 ,300 sine 20 divided by course 30.

01:20 So let me show the value for f x.

01:24 So 1 ,300 into sine 20 divided by cos 30.

01:34 So this is 513 .41 newtons.

01:41 Now let us say what does the second part say? second part says that the normal force what will use a normal force right so normal force along the perpendicular to the plane would be calculated using the considering equating the net force in the perpendicular to the plane equal to zero so n plus 513 .41 cost 30 sorry sign let me see it is sine sign 30 would be equal to 1500.

02:17 Post 20 right so n would be equal to 1500 cost 20 minus 513 .41 into sine 30 so this is coming out to be 11 52 .8 3 newtons...

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