Video: Finding Two Unknown Components in a Force Vector given the Vector of the Moment of the Force about a Point in Three Dimensions

If the force 𝐹, where 𝐹 = βˆ’2𝑖 + 𝐿𝑗 βˆ’ 9π‘˜, is acting on the point 𝐴 (4, 5, βˆ’2), and the moment 𝑀_𝐡 of the force about the point 𝐡 (βˆ’4, βˆ’4, 3) is βˆ’91𝑖 + 82𝑗 + 2π‘˜, determine the value of 𝐿.

03:49

Video Transcript

If the force 𝐹, where 𝐹 equals negative two 𝑖 plus 𝐿 𝑗 minus nine π‘˜, is acting on the point 𝐴 β€” four, five, negative two β€” and the moment 𝑀 sub 𝐡 of the force about the point 𝐡 β€” negative four, negative four, three β€” is negative 91𝑖 plus 82𝑗 plus two π‘˜, determine the value of 𝐿.

Given the force 𝐹 and also given the coordinates of points 𝐴 and 𝐡 and, finally, given the moment 𝑀 sub 𝐡 of the force about the point 𝐡, we want to solve for the value of 𝐿 given in the equation for 𝐹. We’ll start off by drawing in the points 𝐴 and 𝐡 on a set of coordinate axes.

With our 𝑖-, 𝑗-, and π‘˜-directions written in, we can locate points 𝐴 and 𝐡 on this graph. Point 𝐴 has coordinates 𝑖 equals four 𝑗, equals five, and π‘˜ equals negative two. And point 𝐡 has coordinates 𝑖 equals negative four, 𝑗 equals negative four, and π‘˜ equals three. The force 𝐹 acts from point 𝐴. We’re not entirely sure what direction because we don’t know it’s 𝑗-component.

If we connect points 𝐴 and 𝐡 by a vector, we can call that vector 𝑅. It’s the distance vector between the point 𝐴 at which the force acts and the point 𝐡 at which our moment is measured. Based on our diagram, we can say that 𝑅 is equal to 𝐴 minus 𝐡. And when we enter in the coordinates of those two points and calculate 𝑅, we find it’s a vector with components eight 𝑖, nine 𝑗, and negative five π‘˜.

Now we’ll record 𝑅, 𝐹, and 𝑀 sub 𝐡 off to the side so we can begin to solve for 𝐿. Now that we know the displacement vector 𝑅, the force 𝐹, and the moment about point 𝐡, 𝑀 sub 𝐡, we can recall that this moment is equal to the cross product of 𝑅 and 𝐹. This cross product is equal to the determinant of this matrix, with our unit vectors written in followed by the components of our two vectors.

When we apply this relationship to our scenario, we see it’s the 𝐹 sub 𝑗-component we want to solve for. If we know that, then we’ll know 𝐿. We can figure out 𝐹 sub 𝑗 or 𝐿 by focusing either on the 𝑖- or the π‘˜-component of 𝑀 sub 𝐡. Just to choose one of those two ways, let’s focus on the 𝑖-component and let that lead us to the value of 𝐿.

The 𝑖th component of the moment, which we can call 𝑀 sub 𝐡𝑖, is equal to 𝑅 sub 𝑗 sub times 𝐹 sub π‘˜ minus 𝑅 sub π‘˜ times 𝐹 sub 𝑗. And we can replace 𝐹 sub 𝑗 with 𝐿 because that’s what it is. Now the 𝑖th component of 𝑀 sub 𝐡 is negative 91. That’s equal to the 𝑗th component of 𝑅, which is positive nine, times the π‘˜th component of 𝐹, which is negative nine, minus the π‘˜th component of 𝑅, negative five, multiplied by 𝐿.

Negative nine times nine is negative 81. And adding 81 to both sides gives us the expression negative 10 is equal to five 𝐿, or 𝐿 is equal to negative two. That’s the value of the 𝑗th component of 𝐹.

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