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Ch 10: Dynamics of Rotational Motion

Chapter 10, Problem 10

A metal bar is in the xy-plane with one end of the bar at the origin. A force F = 97.00 N)i + (-3.00 N)j is applied to the bar at the point x = 3.00 m, y = 4.00 m. (b) What are the magnitude and direction of the torque with respect to the origin produced by F?

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welcome back everybody. We have a horizontal thin rod that looks like this and it is free to rotate about this point. Oh now this 0.0.0 represents the origin. And we are told that at the coordinate of 1.5 X. And one Y. This is the point five Emma One that a force is applied. Now this force is at a 30° angle. So guy right here data data is 30° and I'm gonna call this force f. or magnitude Of F is equal to 80 Newtons. Now connecting this force to this point, we are going to have a displacement vector given by this arrow. Right here. I'm just gonna deem this arrow are and we are asked to find what the fork is about our point. Oh but we have our N. F. And a formula states that if you take the cross product of these two vectors, you will find a formula for our torque about Oh let's go ahead and do that. So we have that. Our torque equal to r cross F. Well, we need to find R. And F in vector form. So what is our our here while we have this point that starts at the origin. So it's simply just those coordinates put into vector form. We are going to have 1. m in the X. Or I. Unit direction plus one m in the J. Unit direction. Now this is going to be cross or cross multiplied or sorry, the cross product with F. Well, we have our angle measure and we have our magnitude. So we can find the X and Y components of our F. R. X components, but are are in parentheses. So we don't get confused. Our X component is going to be our magnitude times the co sign of our angle. And this is going to be in newtons in the I. Unit direction, plus our magnitude times the sine of our angle in the J. Unit direction. When you take this cross product, we get a final answer of negative 13. newton meters in the A unit direction or the Z axis and negative. It'll be going in the negative Z direction. We have negative 13.3 newton meters. The negative Z direction corresponding to our answer choice of D. Thank you guys so much for watching. Hope this video helped. We will see you all in the next one.
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Textbook Question
A metal bar is in the xy-plane with one end of the bar at the origin. A force F = 97.00 N)i + (-3.00 N)j is applied to the bar at the point x = 3.00 m, y = 4.00 m. (a) In terms of unit vectors i and j, what is the position vector r for the point where the force is applied?
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