Video: Determining the Net Torque about a Point

What is the torque about the origin of the force (5.0𝐒 βˆ’ 2.0𝐣 + 1.0𝐀) N if it is applied at the point whose position is: 𝐫 = (βˆ’2.0𝐒 + 4.0𝐣) m?

07:04

Video Transcript

What is the torque about the origin of the force 5.0𝐒 minus 2.0𝐣 plus 1.0𝐀 newtons if it is applied at the point whose position is 𝐫 equals negative 2.0𝐒 plus 4.0𝐣 meters?

In this problem, we’re given a force that acts at a point, and we’re told the force in terms of its components: 5.0𝐒 minus 2.0𝐣 plus 1.0𝐀 newtons. We can picture this force acting in a three-dimensional space being applied at a point whose distance from the origin negative 2.0𝐒 plus 4.0𝐣 meters.

Because this force has 𝐒, 𝐣, and 𝐀 components, it’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, let’s call that 𝛕, and recognize that it’s a vector. Having both magnitude and direction, we’ll use the Greek letter lower case 𝛕 to represent torque. And let’s begin by recalling the definition of what torque is.

Torque, 𝛕, is defined as the cross product of 𝐫, our position vector and 𝐅, our force vector. Since we’re combining 𝐫 and 𝐅 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 let’s recall that relationship now. The cross product between two three-dimensional vectors, 𝐀 and 𝐁, is equal to these 𝐒, 𝐣, and 𝐀 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 𝐀 and 𝐁, say this first pair 𝐴 sub 𝑦 times 𝐡 sub 𝑧, the next pair that we see is the reverse of that with subscripts 𝐴 sub 𝑧 and 𝐡 sub 𝑦. And this pattern goes on for the 𝐣 and 𝐀 components as well. Also notice that for each one of the three components 𝐒, 𝐣, and 𝐀, the components of the vectors 𝐀 and 𝐁 that are used to calculate that term are perpendicular or orthogonal to those components.

For example, with the 𝐣 component of 𝐀 cross 𝐁, normally associated with the 𝑦-value of those vectors, we see the only components of 𝐀 and 𝐁 involved are π‘₯ and 𝑧 orthogonal to 𝑦. This three-dimensional cross product formula is useful to have memorized, but if not, we can always look it up. Let’s now apply this cross product formula to our scenario using 𝐫 and 𝐅 in place of 𝐀 and 𝐁, respectively. And just a quick note, order is very important with cross products. For example, 𝐫 cross 𝐅 is not the same as 𝐅 cross 𝐫. In general, those two cross products are not equal.

So we want to find the cross product of 𝐫 cross 𝐅 which is equal to the torque. To do that, let’s apply our cross product equation to the 𝐫 vector and 𝐅 vector given to us in the problem statement.

As we work along, 𝐫, our vector for position, is represented by 𝐀 in the cross product formula. And 𝐅, our vector for force, is represented by 𝐁. Let’s begin with the 𝐒th component of 𝐫 cross 𝐅. 𝐫 cross 𝐅 is equal to π‘Ÿ sub 𝑦 times 𝐹 sub 𝑧 minus π‘Ÿ sub 𝑧 times 𝐹 sub 𝑦. For the 𝐣 component of 𝐫 cross 𝐅, we have π‘Ÿ sub 𝑧 times 𝐹 sub π‘₯ minus π‘Ÿ sub π‘₯ times 𝐹 sub 𝑧. And for the 𝐀th component, we have π‘Ÿ sub π‘₯ times 𝐹 sub 𝑦 minus π‘Ÿ sub 𝑦 times 𝐹 sub π‘₯.

We’re now ready to begin to plug in the values for these components of 𝐫 and 𝐅, working off of the 𝐫 and 𝐅 we have defined in the problem statement. Let’s start with the 𝐒th component of 𝐫 cross 𝐅. π‘Ÿ sub 𝑦 is 4.0. 𝐹 sub 𝑧 1.0. π‘Ÿ sub 𝑧 is zero, because there is no 𝐀th component of 𝐫. And 𝐹 sub 𝑦 is negative 2.0. 4.0 times 1.0 minus 0 is 4.0.

Now we move on to the 𝐣th component of 𝐫 cross 𝐅. π‘Ÿ sub 𝑧 is zero, because there is no 𝐀th component of 𝐫. π‘Ÿ sub π‘₯ is negative 2.0. And 𝐹 sub 𝑧 is 1.0. Zero minus negative 2.0 times 1.0 all together equals 2.0. Now moving on to the 𝐀th component, π‘Ÿ sub π‘₯ is negative 2.0, 𝐹 sub 𝑦 is negative 2.0, π‘Ÿ sub 𝑦 is 4.0, and 𝐹 sub π‘₯ is 5.0. Negative 2.0 times negative 2.0 minus 4.0 times 5.0 all together equals negative 16.

Since we’ve solved for 𝐫 cross 𝐅, by the definition of torque we’ve also solved for the torque. We find that the torque in this scenario is equal to 4.0𝐒 plus 2.0𝐣 minus 16𝐀. 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.

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