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# Components of a Force - Problem 2

###### Norm Prokup

###### Norm Prokup

**Cornell University**

PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

We're talking about finding the components of a force, you don't always want to find the horizontal, and vertical components of a force. Here is an example of a problem where you want to find something else.

A 300 pound crate rests on an inclined plane. An inclined plane is basically just a ramp. It says find the components of its weight perpendicular to and parallel to the plane.

So here is our 300 pound crate, this vector represents its weight. The inclined plane has an incline of 20 degrees. So the components of it's weight perpendicular to, and parallel to the plane would look like this. This is the one perpendicular straight into the plane, and of course the one parallel. So let's call the one perpendicular 'a', and the one parallel 'b'. I want to find the magnitudes of a and b.

Now it would really be helpful if I can find some of the angles of this triangle. Now you might suspect that this angle is 20 degrees because it kind of looks like this one, but let's actually show that.

Just really quickly, I can draw component b up here and it's going to be parallel to the surface of the plane. And if you think about this triangle extending this vertical side, this vertical downward, we have a little right triangle here, a 20 degree angle here. And you know that the other acute angle has to be 70, because the two acute angles in a right triangle have to add up to 90. So this angle 70 and that means this angle is 70.

Now if this angle is 70, this angle is also 70. So I'll write 70 degrees, and that means that this one is 20. That's all we need to find the components of this force. So here is a larger version of that same triangle, a 20 degree angle. This one is called 'a', this component is called b, so let's find the magnitudes of a and b.

First 'a', let's observe that 'a' can be found using the cosine of 20 degrees. Cosine of 20 degrees equals the magnitude of 'a' over 300. Now keep in mind it would not be correct to write cosine of 20 degrees equals 'a' over 300, you've got to write the magnitude. The magnitude is a real number, 'a' is a vector.

So we multiply both sides by 300, we get 300 cosine 20 equals the magnitude of a. So let's multiply this 300 cosine 20, and I get 281.9 and this is going to be in Pounds. That's an approximation for the component that's perpendicular to the plane.

What about b? Well for b, I could use the sine of the angle, sine of 20 degrees is the magnitude of b over 300. So the magnitude of b equals 300 times the sine of 20. So again we go to the calculator, second entry and change the cosine to a sine, and I get 102.6, 102.6 pounds.

Now let's take a look at what that means. We have a force of 281.9 pounds acting perpendicular to the plane, that's the normal force of the box against the plane. And then, the component parallel has a magnitude of 102.6 pounds, and that means it takes 102.6 pounds in this direction to keep the crate from sliding. So this force may be provided by friction, or maybe a little man holding it up or a rope tied to the top somehow, but anyway that's basically what's keeping this thing from sliding down.

The normal force is the force of the box pressing into the inclined plane. Again the normal force is 281.9 pounds, the parallel force is 102.6.

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###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

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