Calculate the magnitude of the normal force on a 177 kg...

Calculate the magnitude of the normal force on a 17.7 kg block in the following circumstances. (Enter your answers in N.) (a) The block is resting on a level surface. N (b) The block is resting on a surface tilted up at a 30.8° angle with respect to the horizontal. N (c) The block is resting on the floor of an elevator that is accelerating upward at 3.53 m/s2. N (d) The block is on a level surface and a force of 175 N is exerted on it at an angle of 30.8° below the horizontal. N

Transcript

00:01 Okay, we're looking to find the normal force acting on a block in several different situations.

00:08 So we first draw our picture, we define a coordinate system.

00:14 We draw the forces either on the picture or in a free body diagram.

00:20 Then we apply newton's second law.

00:22 Some of the forces, in this case we'll just do it in the y direction equals ma.

00:28 The normal force minus the weight.

00:31 It's not accelerating, so the right hand side is zero.

00:33 The normal force is equal to ng.

00:39 For our 17 .7 kilogram block this gives us a normal force of 173 .46 newtons.

00:50 Okay, next our block is placed on an inclined plane, inclined at this angle of theta, 30 .8 degrees.

01:04 And what is the normal force? so for these kind of inclined plane block problems, you want to make the x -axis parallel to the slope and the y -axis normal to the slope.

01:20 Let's draw our free body diagram over here to the side.

01:24 M -g normal force.

01:30 Remember this is the y direction.

01:37 So there's our angle theta.

01:40 Some of the forces in the y equals, this is not accelerating, so it equals zero.

01:45 We'd have the normal force minus m g.

01:50 Cosine theta.

01:52 Cosine theta gives you this component.

01:55 Of the force, the y component, that equals zero.

02:00 Okay, so the normal force equals mg cosine theta, which is 148...