Video Transcript
Given that 𝐅 one equals two 𝐢
minus 𝐣, 𝐅 two equals five 𝐢 plus two 𝐣, and 𝐅 three equals negative three 𝐢
plus two 𝐣 are acting on the point 𝐴 one, one, determine the moment of the
resultant of these forces about the two points 𝐵 two, one and 𝐶 six, four. Which of the following would we
conclude about the line of action of the resultant force?
The first step to solving this
problem is to find the resultant of the forces acting on the point 𝐴. We can do this using simple vector
addition of 𝐅 one, 𝐅 two, and 𝐅 three. The resultant force, which we will
call 𝐅 𝑟, is equal to the sum 𝐅 one plus 𝐅 two plus 𝐅 three. So this comes to two 𝐢 minus 𝐣
plus five 𝐢 plus two 𝐣 minus three 𝐢 plus two 𝐣. Adding these component-wise gives
us four 𝐢 plus three 𝐣. So we now have the resultant of the
three forces 𝐅 𝑟 equals four 𝐢 plus three 𝐣.
We now need to find the moment of
this resultant force about the two points 𝐵 and 𝐶, which we can find by taking the
cross product 𝐫 cross 𝐅, where 𝐫 is the vector from the pivot point 𝑂 to the
point of action 𝑃. For the moment about the point 𝐵,
the vector 𝐫 is the vector 𝐵 to 𝐴, which is given by one, one minus two, one,
which is equal to negative one, zero.
Therefore, the moment about the
point 𝐵 is given by the determinant of the three-by-three matrix 𝑖, 𝑗, 𝑘,
negative one, zero, zero, four, three, zero. Both 𝑟 and 𝐹 are in the
𝑥𝑦-plane with a 𝑘-component of zero. Therefore, only the 𝑘-component of
their cross product will be nonzero.
Taking this determinant by
expanding along the top row gives us negative three 𝑘. Let’s move this up here and clear a
little space.
Now, for the moment about the point
𝐶, the vector 𝐫 is equal to the vector 𝐂𝚨, which is given by one, one minus six,
four, which is equal to negative five, negative three. Therefore, the moment 𝑀 𝑐 of the
resultant force about the point 𝐶 is given by the determinant of the three-by-three
matrix 𝑖, 𝑗, 𝑘, negative five, negative three, zero, four, three, zero.
Once again, both vectors are in the
𝑥𝑦-plane and have a 𝑘-component of zero. Therefore, only the 𝑘-component of
their cross product will be nonzero. Taking this determinant by
expanding along the top row gives us once again negative three 𝑘.
We therefore have the first part of
our answer. The moment of the resultant of the
forces about the point 𝐵 is equal to negative three 𝑘 and about the point 𝐶 is
also equal to negative three 𝑘.
Now, we need to decide which of the
following options we can conclude about the line of action of the resultant
force. To begin with, we know that all of
the points involved are in the 𝑥𝑦-plane and all of the forces involved have no
𝑘-component. Therefore, we only need to consider
the problem in two dimensions. We have two points 𝐵 and 𝐶, and
the moment of the same force 𝐹 𝑟 equals four 𝑖 plus three 𝑗 is equal to negative
three 𝑘 in both cases. It has the same magnitude and the
same direction.
In both cases, the moment is
directed into the screen. So, from this perspective, it is a
clockwise-turning force. The line of action of the force
must firstly be the same perpendicular distance from both points and must also be on
the same side as both points, meaning that it does not pass between them.
We can therefore conclude answer
(B) that the line of action of the resultant force is parallel to the line 𝐵𝐶.
