Question Video: Calculating the Acceleration of an Object under Two Forces

Video Transcript

A particle is accelerated when acted on by two forces, as shown in the accompanying diagram. The force 𝐹 sub 𝐵 has twice the magnitude of force 𝐹 sub 𝐴. Find the direction in which the particle’s net acceleration occurs in terms of the angle below the negative 𝑥-direction from the position of the particle.

So given the two forces acting on the particle, 𝐹 sub 𝐵 and 𝐹 sub 𝐴, we want to solve for a direction, the direction in which the particle’s net acceleration occurs. We’ll call this direction 𝜃, where 𝜃 is an angle measured from the negative 𝑥-direction from the position of the particle. In order to solve for the direction of the particle’s acceleration, we’ll want to solve for the resultant force acting on the mass 𝑚. If we call that resultant force 𝐹 sub 𝑅, we know that force will have two components. One we can call 𝐹 sub 𝑥 in the 𝑖-direction, that is, along the 𝑥-axis. And the second component we can call 𝐹 sub 𝑦 in the 𝑗-direction along the 𝑦-axis.

To solve for 𝐹 sub 𝑅, the resultant force, we’ll add together 𝐹 sub 𝐵 and 𝐹 sub 𝐴, the two forces acting on the mass 𝑚. We can start by writing the force vector 𝐹 sub 𝐴 as the magnitude of that force multiplied by its direction in the 𝑖-direction. Likewise, we can write the force vector 𝐹 sub 𝐵 as negative the magnitude of that force multiplied by the cos of 45 degrees — that’s the 𝑖-component of 𝐹 sub 𝐵 — minus the magnitude 𝐹 sub 𝐵 times the sin of 45 degrees — that’s the 𝑗-component of that force.

In the exercise statement, we’re told something about the magnitude 𝐹 sub 𝐵. We’re told that it’s equal to twice the magnitude of 𝐹 sub 𝐴. We substitute in two 𝐹 sub 𝐴 for 𝐹 sub 𝐵 and factor it out along with the minus sign from this expression for the force vector 𝐹 sub 𝐵. We can further simplify this expression for 𝐹 sub 𝐵 by realizing that the cos of 45 degrees and the sin of 45 degrees are both the square root of two over two. And we see that now the factors of two cancel out, so that 𝐹 sub 𝐵, the vector, is equal to negative root two the magnitude of 𝐹 sub 𝐴 in the 𝑖- and the 𝑗-direction.

Our next step is to add these two forces, 𝐹 sub 𝐴 and 𝐹 sub 𝐵, together. We find that their sum, which is the resultant force 𝐹 sub 𝑅, is equal to the magnitude of 𝐹 sub 𝐴 times one minus the square root of two 𝑖 minus the square root of two 𝑗. We’ve now solved for the resultant or net force acting on our particle. And if we were to sketch in approximately the direction of this force on our diagram, it might point in a direction like this, slightly farther towards the negative 𝑦-direction and 𝐹 sub 𝐵. The particular direction this force and therefore the particle’s acceleration is directed is given by the angle 𝜃 measured from the negative 𝑥-axis with respect to the particle’s position. If we were to draw in the 𝑥- and 𝑦-components of that resultant force, 𝐹 sub 𝑅, on our diagram, we can see that the angle 𝜃 is part of that right triangle. And we can see that the tangent of that angle 𝜃 is equal to the 𝑦-component of the resultant force 𝐹 sub 𝑦 divided by the 𝑥-component 𝐹 sub 𝑥.

Looking at the resultant force we solved for earlier, we see that the 𝑦-component of that force is the magnitude 𝐹 sub 𝐴 times negative root two. And we see further that the 𝑥-component of that force is the magnitude of 𝐹 sub 𝐴 multiplied by one minus the square root of two. Looking at this fraction, we see that the magnitudes 𝐹 sub 𝐴 cancel out. And if we then take the inverse tangent of both sides of the equation, we find that 𝜃 is equal to the inverse tangent of negative the square root of two divided by one minus the square root of two. Entering this expression on our calculator, we find that, to two significant figures, 𝜃 is 74 degrees. That’s the direction of the particle’s acceleration as measured from the negative 𝑥-axis with respect to the particle’s position.