Four dogs named Ang, Bing, Chang, and Dong play a game of tug of war with a toy. Ang pulls on the toy in a direction 55.0 degrees south of east, Bing in a direction 60.0 degrees east of north, and Chang in a direction 55.0 degrees west of north. Ang pulls strongly with a force of magnitude 0.160 kilonewtons. Bing pulls even more strongly than Ang with a force of magnitude 0.200 kilonewtons, and Chang pulls with a force of magnitude 0.140 kilonewtons. When Dong pulls on the toy in such a way that his force is equal in magnitude and opposite in direction to the resultant of the other three forces, the toy does not move in any direction. What magnitude force does Dong pull on the toy with? At what angle south of west does Dong pull on the toy?
In part one of this exercise, we want to solve for the magnitude of the force with which Dong pulls on the toy. We’ll call that magnitude 𝐹 sub 𝐷. And in part two, we want to solve for the angle south of west at which Dong pulls on the toy. We’ll name that angle 𝜃. Let’s start out by drawing a diagram showing the way that Ang, Bing, and Chang are pulling on the toy.
If we draw the four compass directions: north, south, east, and west on a grid, then we can place the toy that the four dogs are pulling on at the center of those axes. We’re told that the dog Ang pulls on the toy at an angle 55.0 degrees south of east. We can call that force vector 𝐹 sub 𝐴 and the magnitude of that force is given as 160 newtons or 0.160 kilonewtons.
We’re also told that the dog Bing pulls on the toy at an angle 60.0 degrees east of north. And if we call that force vector 𝐹 sub 𝐵, we know the magnitude of that vector is 0.200 kilonewtons. We also know that the dog Chang pulls on the toy at an angle 55.0 degrees west of north. And if we call that force vector 𝐹 sub 𝐶, we know its magnitude is 0.140 kilonewtons. The fourth dog Dong also pulls on the toy. We can call that force vector 𝐹 sub 𝐷. And the angle south of west at which Dong pulls we’ve called 𝜃.
To solve for the magnitude of that force 𝐹 sub 𝐷, we can recall the fact that the toy is in equilibrium; that is, it’s not moving. That means that if we add up the forces of Ang, Bing, and Chang and take their magnitude, then that result must be equal to the magnitude with which Dong pulls 𝐹 sub 𝐷. To calculate that magnitude, we can start out by writing out the components of 𝐹 sub 𝐴.
On our diagram, we’ll let east be the direction that defines the direction of the 𝑖 unit vector and we’ll let north define the direction of the 𝑗 unit vector so that each of our four vectors can be written in components of 𝑖 and 𝑗. Using this convention, we can write the force with which Ang pulls as the magnitude of that force 160 newtons times the cosine of 55.0 degrees in the 𝑖 direction minus the sine of 55.0 degrees in the 𝑗 direction.
We can follow a similar process for the force with which Bing pulls — writing the force as its magnitude multiplied by its 𝑖 component plus its 𝑗 component — and also for 𝐹 sub 𝐶, the force with which the dog Chang pulls. Now that we have 𝐹 sub 𝐴, 𝐹 sub 𝐵, and 𝐹 sub 𝐶 written according to their components, we want to add the three together. When we do, being careful to add by columns in the 𝑖 and 𝑗 components separately, we find a result of 150.3𝑖 plus 49.24𝑗 newtons.
This result is significant because we’re told in the problem statement that 𝐹 sub 𝐷, the force with which Dong pulls, is equal and opposite to the sum of the forces of the other three dogs. This means that if we insert a minus sign in front of our force vector, then we’ve solved for 𝐹 sub 𝐷 — the vector of the force with which Dong pulls.
This is a good result to keep in mind. But for part one, we want to solve for the force magnitude. To find the magnitude, we’ll take the square root of the 𝑖 component of 𝐹 sub 𝐷 squared plus the 𝑗 component of 𝐹 sub 𝐷 squared. When we plug these components in to our square root equation and calculate this term, we find that to three significant figures 𝐹 sub 𝐷 is 158 newtons. That’s the magnitude of the force which with the dog Dong pulls on the toy.
In part two, we want to solve for 𝜃, where 𝜃 is the angle South of West at which Dong pulls on the toy. If we look at the triangle that’s created by the force magnitude 𝐹 sub 𝐷, the angle 𝜃, and the horizontal axis, we see that the triangle is a right triangle with sides 𝐹 sub 𝐷𝑖, the 𝑖 component of 𝐹 sub 𝐷, and the 𝑗 component of 𝐹 sub 𝐷 — 𝐹 sub 𝐷𝑗 — with the hypotenuse of the magnitude of the force 𝐹 sub 𝐷.
All of this means that we can write the tangent of the angle 𝜃 as 𝐹 sub 𝐷𝑗 divided by 𝐹 sub 𝐷𝑖 according to the geometry of this triangle. We’ve solved for those components of 𝐹 sub 𝐷 — the 𝑖𝐹 component in the 𝑗𝐹 component — and can plug those into this equation now.
With those components added in, if we then take the arc tangent of both sides of our equation, then we now have an expression for that angle 𝜃 we want to solve for. When we enter this value on our calculator, we find that to three significant figures 𝜃 is 18.1 degrees. That’s the direction south of west at which Dong pulls on the toy.