PHYS 201 SD1 FD4 Solution S2026

Añadido: 2026-05-12

[00:00:00]Okay, we're here to talk about today's quizzes. Um, we'll start with FD4A.

[00:00:06]FD4A and FD4B were the same. So, we've got the pulley system like what we did last week. Label the force arrows in the following free body diagrams. So, for A, you've got tension pulling to the right, normal force, Fgraph, FK, friction.

[00:00:25]For box B, you had tension pulling up and F grab pulling down. That was it.

[00:00:31]Part B, which forces are Newton's third law pairs? Well, the tensions are the same. So, T equals T. Also, for box A, the normal force counteracts the gravity force. So, these two are a Newton Newton's third law pair.

[00:00:48]Okay, let's do ST1A. A little bit complicated. So we have 8 kg mass that's being pulled with a tension. I tell you that the coefficient of kinetic friction is 0.15. What's the net acceleration of the block? Let me zoom in a little bit.

[00:01:02]Okay. So let's start with the free body diagram. Free body diagram. You got normal F graph friction and tension which has an X and Y component. We're going to start by looking at all the vectors in the Y direction. Fnet Y equals normal force plus TY minus FG.

[00:01:20]and that equals zero. It's not moving up and down. We're going to isolate solve for normal. So normal equals fg minus t y. fg is mg. ty is t sin theta. So the normal force is after you plug in all the numbers 8 * 9.8 - 35 * sin 25.

[00:01:41]Normal force is 63.608, right? 63.6.

[00:01:47]Once you know the normal force, we can figure out the friction. So here if we look at fnet in the x direction, fnet will equals m which equals tx - fk. tx and fk are the only two things in the x direction. M= t cosine thetus mu k normal force. I want to isolate for a net. So I divide both sides by m. A= t cosine theta - mu k / n / m. Plug in numbers. T cosine theta is 35 cosine 25.

[00:02:21]Mu k is given as 0.15. The normal force we calculated over here from the y vectors divided by the mass. A net is 2.77 m/s squared.

[00:02:35]Okay.

[00:02:37]SD1B was slightly different.

[00:02:43]So similar situation except that I say that the net acceleration is 2 m/s squared. What is mu k? All right, same free body diagram. We're going to do the y vectors. fnet y is normal force plus ty minus fg that equals zero. Going to solve the normal force. That's fg - t y.

[00:03:04]That's mgus t sin theta. I plug in the numbers. My normal force is 81.095. 095 newtons. Okay, once I know normal force, I can go in and look at the X vectors.

[00:03:18]Fnet X= MAT= TX minus FK. So we have MAT= T cosine theta minus MK normal. I'm going to isolate from Muk normal.

[00:03:28]Isolate for FK basically. So if I move that to our side, muk * normal equals T cosine theta minus MA. Divide both sides by normal force. And so now if I plug in numbers, mu k is 40 cosine 25 - 10 * 2 / 81.095.

[00:03:48]Muk k will come out to be 0.2.

[00:03:52]That was it. If you have any questions about this, please reach out. Let me know.