What is the torque about the origin
of the force 5.0𝐢 minus 2.0𝐣 plus 1.0𝐤 newtons if it is applied at the point
whose position is 𝐫 equals negative 2.0𝐢 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:
5.0𝐢 minus 2.0𝐣 plus 1.0𝐤 newtons. We can picture this force acting in
a three-dimensional space being applied at a point whose distance from the origin
negative 2.0𝐢 plus 4.0𝐣 meters.
Because this force has 𝐢, 𝐣, and 𝐤 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 𝐢, 𝐣, and 𝐤 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 𝐵
sub 𝑧, the next pair that we see is the reverse of that with subscripts 𝐴 sub 𝑧
and 𝐵 sub 𝑦. And this pattern goes on for the 𝐣
and 𝐤 components as well. Also notice that for each one of
the three components 𝐢, 𝐣, and 𝐤, the components of the vectors 𝐀 and 𝐁 that
are used to calculate that term are perpendicular or orthogonal to those
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 𝑥 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
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 𝐢th component of 𝐫 cross 𝐅. 𝐫 cross 𝐅 is equal to 𝑟 sub 𝑦
times 𝐹 sub 𝑧 minus 𝑟 sub 𝑧 times 𝐹 sub 𝑦. For the 𝐣 component of 𝐫 cross 𝐅,
we have 𝑟 sub 𝑧 times 𝐹 sub 𝑥 minus 𝑟 sub 𝑥 times 𝐹 sub 𝑧. And for the 𝐤th component, we
have 𝑟 sub 𝑥 times 𝐹 sub 𝑦 minus 𝑟 sub 𝑦 times 𝐹 sub 𝑥.
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 𝐢th
component of 𝐫 cross 𝐅. 𝑟 sub 𝑦 is 4.0. 𝐹 sub 𝑧 1.0. 𝑟 sub 𝑧 is zero, because there is
no 𝐤th 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. 𝑟 sub 𝑧 is zero, because there is
no 𝐤th component of 𝐫. 𝑟 sub 𝑥 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 𝐤th
component, 𝑟 sub 𝑥 is negative 2.0, 𝐹 sub 𝑦 is negative 2.0, 𝑟 sub 𝑦 is 4.0,
and 𝐹 sub 𝑥 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 4.0𝐢 plus 2.0𝐣 minus 16𝐤. 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.