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# Torque

###### Matt Jones

###### Matt Jones

**M.Ed., George Washington University**

Dept. chair at a high school

Matt is currently the department chair at a high school in San Francisco. In his spare time, Matt enjoys spending time outdoors with his wife and two kids.

**Torque**, also called movement of force, measures how much a force rotates an object around an axis. **Torque** is the cross product of the force vector and the vector of the distance from the axis. The formula for *rotational torque = radius x force x sin*(angle between force and radius)

Torque you've probably experienced torque if you've tried to loosen a bolt with a really small wrench and you apply a lot of force to it and you just can't get that bolt lose alright it's just too tough. Well if you can get a longer wrench you can apply that same force and you can get that bolt off right, because a torque is a force but it's also dependent on the distance from the center of the axis right so the definition of force that makes an object turn around an axis, but remember here that force is dependent on the distance from the axis right. So the unit we use for torque is Newtons per meter okay.

Now let's look at how we could solve for that rotational torque that force that we use to move things around an axis okay. So the formula we're going to use for that torque this is actually a Greek t called a tao sometimes it's written as a regular t but torque is equal to the radius that's the distance from the center times the force times the sine of the force that's applied. So let's look at a couple of problems and try to figure out how can we take the problems in a homework question and apply them into this equation for torque okay. So let's say I'm opening a door okay, well a door that's not a circle, well it is part of a circle right when I open that door it's revolving around an axis a hedge okay.

Let's say I'm applying 15 Newtons to open the door and when I apply that I'm applying it directly perpendicular to the hedge. So the angle what would be my angle there? That would be 90 degrees. So that's going to be the angle of the force relative to the central axis okay and what's the radius? Well a door, does a door have a radius? Well it does it's the distance of the door, so that would be my radius. So if I plug those in and I say torque equals radius times Newtons times sine theta okay. I'll just plug those in I've got 1 meter times 50 Newtons and the sine of 90 degrees is 1 okay.

Now if I saw that I just get my torque is 50 Newtons meters, pretty easy pretty straight forward problem once I figure out how to apply all of these values from my equation into there. Let's look at one that might be a little bit more challenging. Now let's say we've got a merry-go-round and we're pushing a small merry-go-round the radius of that merry-go-round is one and a half meters and I'm pushing it at an angle, I'm not pushing it tangentially I'm pushing it at an angle that if I calculate the force angle with the center to the axis I get an angle of 110 degrees okay. Well again I have my radius so if I say torque equals radius times Newtons times the sine of theta again I can just put in my numbers and I've got 1.5 from my radius I've got 50 Newtons and the sine of this 110 is going to be 0.94 so if I, 1.5 times 50 is 75 times 0.94 is 70.4 Newton, I forgot my meters there right so let's go ahead and add that and again this is quite common where you kind of work through a problem and then you forget a unit so just put it back in there, Newton meter and that Newton meters is the unit of torque that's applied on the merry-go-round. So these are 2 different types of problems that you'd have and you can use in this formula once you can figure out the way things go into that formula to solve for torque.

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###### Matt Jones

M.Ed., George Washington University

Matt is very comfortable in front of the whiteboard and is able to make every topic easy for anyone to digest. His straightforward approach to teaching is very refreshing.

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