FIGURE P3.43 shows three ropes tied together in a knot. One of your friends pulls on a rope with 3.0 units of force and another pulls on a second rope with 5.0 units of force. How hard and in what direction must you pull on the third rope to keep the knot from moving? Give the direction as an angle below the negative x-axis.

Video transcript

Hey, everyone. So this problem is working with force vectors. Let's see what they are giving us and what they're asking us. So we have a physics instructor directing three students to pull a ring. And the ring here is placed at the origin. So we've got three students with different abilities. Therefore, they're exerting different forces on the ring student. One applies a four Newton force. And we can see from this figure that it is at 60 degrees above the X axis student two is exerting a six student force and that's just in the negative X direction. And then we need to determine the force that the third student student three must apply to keep the ring stationary, were asked to express that direction as an angle measured from the X axis. So the keys here are to recognize that if the ring is stationary, that means that the sum of all of these forces must be zero. So we can write that as some of F equals zero. And that's going to be F one plus F two plus F three. Now we're solving for F three, the force of the third student. And that is going to equal uh minus F one minus F two Because they are all equal zero. So we're just rearranging that equation there. And that's how we're going to solve for F three. So first for F one and F two, we're going to rewrite those in vector form. So F one Is going to be four newtons and the x component of that is going to be the co sign of and that's in the I direction that's positive. And then the J component or the Y component is going to be four times the sine of 60. And again, that's also going to be the positive Y. So when we plug that into our calculators, we get to I Plus 3.46 J newts for F two, it's a little bit easier because we only have one component. There's no Y component here and we can recognize that this is in the negative X direction. So it's going to be negative six newtons in the, in the I direction. So when we go to add the F1 and F2 components together to get our F3, we'll just plug those in and that looks like, so minus F one, so it's gonna be minus two, I minus 3. J and then minus F two minus a negative and a positive plus six I again, we're still working in units of Newton. So when we add the I components, we get positive for I -3.46 J. So that's the first part of our answer. So if we look at our possible solutions, we can see that C and D are incorrect off the bat. And so now all we need to do is figure out the angle that that force is being applied in. And so when we have the X and the Y components, we can recall from our trigonometry that we're gonna be using the tangent. So tangent of theta is going to be our Y component over our X component. It's gonna be negative 3.46 divided by four. So the, the data is going to be tangent inverse tangent of that 3.46, negative 3.46 over four plug that in And we get 40.9°. So that's 40.9°. We know with a negative why component and a positive J component, negative Y component and positive X component. Sorry. Um We are looking in the fourth quadrant so we are 40.9 degrees below the positive X axis. So we're right here. That's gonna be 40.9°. And so we look at our potential answers and that is answer B It is four, I -3.46 J Newtons applied at an angle of 40.9° below the positive x axis. That's all we have for this problem. We'll see you in the next video