# Video: Determining the Net Torque about a Point

What is the torque about the origin of the force (5.0π’ β 2.0π£ + 1.0π€) N if it is applied at the point whose position is: π« = (β2.0π’ + 4.0π£) m?

07:04

### Video Transcript

What is the torque about the origin of the force 5.0π’ minus 2.0π£ plus 1.0π€ newtons if it is applied at the point whose position is π« equals negative 2.0π’ plus 4.0π£ meters?

In this problem, weβre given a force that acts at a point, and weβre told the force in terms of its components: 5.0π’ minus 2.0π£ plus 1.0π€ newtons. We can picture this force acting in a three-dimensional space being applied at a point whose distance from the origin negative 2.0π’ plus 4.0π£ meters.

Because this force has π’, π£, and π€ components, itβs a three-dimensional force. And we expect that our resulting torque will also be in three dimensions. We want to solve for the torque, letβs call that π, and recognize that itβs a vector. Having both magnitude and direction, weβll use the Greek letter lower case π to represent torque. And letβs begin by recalling the definition of what torque is.

Torque, π, is defined as the cross product of π«, our position vector and π, our force vector. Since weβre combining π« and π through a cross product, we know that the result, torque, will itself be a vector having magnitude and direction. It will be important for us to recall the formula that teaches us how to compute a cross product between two vectors, so letβs recall that relationship now. The cross product between two three-dimensional vectors, π and π, is equal to these π’, π£, and π€ components that we see written here. This equation may look long and confusing, but there is an order and a method to it.

First of all, notice that for every pair of components of the vectors π and π, say this first pair π΄ sub π¦ times π΅ sub π§, the next pair that we see is the reverse of that with subscripts π΄ sub π§ and π΅ sub π¦. And this pattern goes on for the π£ and π€ components as well. Also notice that for each one of the three components π’, π£, and π€, the components of the vectors π and π that are used to calculate that term are perpendicular or orthogonal to those components.

For example, with the π£ component of π cross π, normally associated with the π¦-value of those vectors, we see the only components of π and π involved are π₯ and π§ orthogonal to π¦. This three-dimensional cross product formula is useful to have memorized, but if not, we can always look it up. Letβs now apply this cross product formula to our scenario using π« and π in place of π and π, respectively. And just a quick note, order is very important with cross products. For example, π« cross π is not the same as π cross π«. In general, those two cross products are not equal.

So we want to find the cross product of π« cross π which is equal to the torque. To do that, letβs apply our cross product equation to the π« vector and π vector given to us in the problem statement.

As we work along, π«, our vector for position, is represented by π in the cross product formula. And π, our vector for force, is represented by π. Letβs begin with the π’th component of π« cross π. π« cross π is equal to π sub π¦ times πΉ sub π§ minus π sub π§ times πΉ sub π¦. For the π£ component of π« cross π, we have π sub π§ times πΉ sub π₯ minus π sub π₯ times πΉ sub π§. And for the π€th component, we have π sub π₯ times πΉ sub π¦ minus π sub π¦ times πΉ sub π₯.

Weβre now ready to begin to plug in the values for these components of π« and π, working off of the π« and π we have defined in the problem statement. Letβs start with the π’th component of π« cross π. π sub π¦ is 4.0. πΉ sub π§ 1.0. π sub π§ is zero, because there is no π€th component of π«. And πΉ sub π¦ is negative 2.0. 4.0 times 1.0 minus 0 is 4.0.

Now we move on to the π£th component of π« cross π. π sub π§ is zero, because there is no π€th component of π«. π sub π₯ is negative 2.0. And πΉ sub π§ is 1.0. Zero minus negative 2.0 times 1.0 all together equals 2.0. Now moving on to the π€th component, π sub π₯ is negative 2.0, πΉ sub π¦ is negative 2.0, π sub π¦ is 4.0, and πΉ sub π₯ is 5.0. Negative 2.0 times negative 2.0 minus 4.0 times 5.0 all together equals negative 16.

Since weβve solved for π« cross π, by the definition of torque weβve also solved for the torque. We find that the torque in this scenario is equal to 4.0π’ plus 2.0π£ minus 16π€. Since we combine units of force that is Newtons and units of distance, that is meters to create torque, the units of torque are Newton meters.