Question Video: Finding the Impulse on a Body Where Three Forces in Vector Form Are Acting on It for a Given Time Mathematics

Three forces, 𝐅₁ = (βˆ’5𝐒 βˆ’ 2𝐣 + 2𝐀) N, 𝐅₂ = (𝐣 βˆ’ 3𝐀) N, and 𝐅₃ = (βˆ’π’ βˆ’ 5𝐣 βˆ’ 2𝐀) N, where 𝐒, 𝐣, and 𝐀 are three mutually perpendicular unit vectors, acted on a body for 3 seconds. Find the magnitude of their combined impulse on the body.

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Video Transcript

Three forces β€” 𝐅 one equals negative five 𝐒 minus two 𝐣 plus two 𝐀 newtons, 𝐅 two equals 𝐣 minus three 𝐀 newtons, and 𝐅 three equals negative 𝐒 minus five 𝐣 minus two 𝐀 newtons, where 𝐒, 𝐣, and 𝐀 are three mutually perpendicular unit vectors β€” acted on a body for three seconds. Find the magnitude of their combined impulse on the body.

In this exercise, we have some body, and we’re told it’s being acted on at the same time by three forces: 𝐅 one, 𝐅 two, and 𝐅 three. These forces are all constant, and we’re told that they act over a time interval we’ll call Δ𝑑 of three seconds. We want to know the magnitude of the impulse on the body from these three forces acting together.

We could calculate the impulse due to each of the three forces individually and then add those impulses together. But a shorter way is to add our three forces together as vectors to solve for the net force acting on our body. So we write out 𝐅 one, 𝐅 two, and 𝐅 three, lining them up by their 𝐒-, 𝐣-, and 𝐀-components. We then add them by their components, starting with the 𝐒-component. Negative five 𝐒 minus 𝐒 gives a total of negative six 𝐒. Then negative two 𝐣 plus 𝐣 minus five 𝐣 adds up to negative six 𝐣. And lastly, two 𝐀 minus three 𝐀 minus two 𝐀 equals negative three 𝐀. The units of these components are all newtons. And this vector overall is equal to the net force acting on our body.

Knowing this, let’s recall a mathematical relationship for the impulse experienced by an object due to a force acting on it. Impulse 𝐒 is equal to 𝐅 times Δ𝑑. In other words, the impulse an object experiences equals the net force acting on that object times the time interval over which that force acts. In our case, we know 𝐅 sub net and we also know Δ𝑑. That was given as three seconds.

But let’s recall that we want to solve for impulse magnitude. That means we’ll need to calculate the magnitude of our net force and use that in our equation. Clearing a bit of space, we can recall that given a three-dimensional vector β€” we’ll call this example vector 𝐯 β€” the magnitude of 𝐯 is equal to the square root of its 𝐒-component squared plus its 𝐣-component squared plus its 𝐀-component squared.

Applying this relationship to our net force, we see it’s equal to the square root of negative six squared plus negative six squared plus negative three squared newtons. This is equal to the square root of 36 plus 36 plus nine newtons or the square root of 81 newtons, which is nine newtons. That’s the magnitude of the net force acting on our body. To calculate 𝐼 then, we substitute this value in for the magnitude of 𝐅 sub net and Δ𝑑 is equal to three seconds. So when we calculate 𝐼, we find a result of 27 newton seconds. That’s the magnitude of the total impulse acting on our body.

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