Video: Finding the Unknown Components of a Group of Forces Acting on a Body Moving Uniformly

A body is moving in a straight line at a constant velocity under the action of a system of forces 𝐅₁, 𝐅₂, and 𝐅₃. If 𝐅₁ = π‘Žπ’ βˆ’ 𝐣 βˆ’ 5𝐀, 𝐅₂ = βˆ’4𝐒 + b𝐣 βˆ’ 3𝐀, and 𝐅₃ = 𝐒 + 6𝐣 + 𝑐𝐀, find π‘Ž, 𝑏, and 𝑐.

02:05

Video Transcript

A body is moving in a straight line at a constant velocity under the action of a system of forces 𝐅 one, 𝐅 two, and 𝐅 three. If 𝐅 one is π‘Žπ’ minus 𝐣 minus five 𝐀, 𝐅 two is minus four 𝐒 plus 𝑏𝐣 minus three 𝐀, and 𝐅 three is 𝐒 plus six 𝐣 plus 𝑐𝐀, find π‘Ž, 𝑏, and 𝑐.

The biggest hint we have here how to answer this question is if we spot that the body is moving at a constant velocity. This indicates what’s that we’re going to need to reference, Newton’s first law of motion. This says that an object either remains at rest or continues to move at a constant velocity unless acted upon by another force.

Another way to think about this is to say that if the net force is zero β€” that is, the vector sum of all forces acting on an object is zero β€” then the velocity of the object is constant. Well, the velocity of our body is indeed constant. So the net force, the vector sum of all forces acting on this body, must be equal to zero. That is, 𝐅 one plus 𝐅 two plus 𝐅 three equals zero.

Let’s replace each of these forces with the given vectors. Now, I’m going to write each of these vectors in column vector form just because I think it makes the next step a little bit easier. In column vector form, π‘Žπ’ minus 𝐣 minus five 𝐀 is π‘Ž, negative one, negative five. The vector 𝐅 two is written as negative four, 𝑏, negative three. And 𝐅 three is then one, six, 𝑐. The sum of these is zero. Now, we could write that as a column vector as zero, zero, zero.

And then, we know that, for the sum of the vectors on the left-hand side to be equal to the vector on the right, their individual components must be equal. If we look at the first component in each vector, that’s π‘Ž, negative four, and one, we can see that π‘Ž minus four plus one must be equal to zero. That is, π‘Ž minus three equals zero. And solving for π‘Ž, we see that π‘Ž must be equal to three.

Similarly, looking at the 𝐣-components, we get negative one plus 𝑏 plus six equals zero or 𝑏 plus five equals zero. And we solve this equation for 𝑏 by subtracting five from both sides. Finally, we repeat this process for the 𝐀-component. We get negative five minus three plus 𝑐 equals zero or negative eight plus 𝑐 equals zero. And we then solve for 𝑐 by adding eight to both sides.

π‘Ž is three, 𝑏 is negative five, and 𝑐 is eight.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.