Question Video: Finding the Magnitude of the Impulse of Two Forces Acting on a Body in Vector Form Mathematics

Two forces 𝐅₁ and 𝐅₂ act on a body of unit mass for 19 seconds. Given that 𝐅₁ = (βˆ’8𝐒 βˆ’ 4𝐣) N and 𝐅₂ = (𝐒 + 3𝐣) N, find the magnitude of the impulse.


Video Transcript

Two forces 𝐅 one and 𝐅 two act on a body of unit mass for 19 seconds. Given that 𝐅 one equals negative eight 𝐒 minus four 𝐣 newtons and 𝐅 two equals 𝐒 plus three 𝐣 newtons, find the magnitude of the impulse.

So in this question, we’re dealing with a body of unit mass. This just means it has a mass of one kilogram. We’re told that two forces, 𝐅 one and 𝐅 two, are acting on this body. And we can add these force vectors to our diagram like this. We’re told that both of these forces act on the body for 19 seconds, and we’re being asked to find the magnitude of the impulse. We can recall that an impulse is effectively a change in momentum. And this can be produced by a force acting on an object for any amount of time. One equation that’s often used to relate force, time, and impulse is this: 𝐒 equals 𝐅 times Δ𝑑. In other words, the impulse produced when a force acts on an object is equal to the force multiplied by the amount of time for which it acts.

In this question, the forces acting on the object have been expressed in two-dimensional vector notation. However, we can still use this equation to find the answer to this problem. Recall that both impulse and force are vector quantities. So really, this equation could be better written if we used vector notation, drawing half arrows over the top of these vector quantities. This equation can now tell us the impulse vector that’s produced when a force vector acts on an object for a given amount of time. The only problem remaining is that in this question, we don’t just have a force; we have two forces acting on the body. However, we can overcome this by just adding the two forces together to find the net force.

We could write 𝐅 net is equal to 𝐅 one plus 𝐅 two. This net force is equivalent to both of these forces combined. 𝐅 one is equal to negative eight 𝐒 minus four 𝐣, and we’re adding 𝐅 two, which is equal to 𝐒 plus three 𝐣. Looking at the 𝐒-terms first, we can see we have negative eight 𝐒 plus 𝐒, which is negative seven 𝐒 in total. And looking at the 𝐣-terms, we have negative four 𝐣 plus three 𝐣, which is negative 𝐣 in total. So now, instead of thinking about two individual forces, 𝐅 one and 𝐅 two, we can just think about the equivalent net force equal to negative seven 𝐒 minus 𝐣.

So now, to calculate the size of the impulse produced by this force, we need to multiply it by Δ𝑑, which is the amount of time for which it acts on the body. We’re told in the question that this is 19 seconds. So the overall impulse vector is equal to our net force vector, that’s negative seven 𝐒 minus 𝐣, multiplied by 19. Here, we’re multiplying a vector quantity force by a scalar quantity time. And we can do this just by multiplying across the parentheses. So we have 19 times negative seven 𝐒, which is equal to negative 133𝐒. And then we have 19 times negative 𝐣, which is, of course, negative 19𝐣. This is the impulse that’s produced by the two forces 𝐅 one and 𝐅 two.

All that’s left to do now is find the magnitude of this impulse vector. To find the magnitude of a vector, we effectively use a form of the Pythagorean theorem. The magnitude of the impulse vector squared is equal to the sum of the squares of its components. So this means that the magnitude itself is equal to the square root of the sum of the squares of each component. In this case, the π‘₯-component is negative 133 and the 𝑦-component is negative 19. Negative 133 squared is 17689, and negative 19 squared is 361. 17699 plus 361 is 18050, and the square root of 18050 is 95 root two.

This value takes the standard units for impulse or momentum. These can be expressed as kilogram meters per second or newton seconds. So there is our final answer. If two forces, one equal to negative eight 𝐒 minus four 𝐣 and the other equal to 𝐒 plus three 𝐣, act on a body of unit mass for 19 seconds, the magnitude of the impulse is 95 root two newton seconds.

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