7. A uniform rod of mass m is placed as shown, with one of...

7. A uniform rod of mass m is placed as shown, with one of its ends resting on a smooth wall while ¼ of the rod's length is sticking out of a rough table. Find the net force the table exerts on the rod at the corner. ? A painting is hung using a cord of length $b$, which is attached to a wall, as shown. The height of the painting is $a$. What is the minimum value of the coefficient of static friction $\mu$, between the wall and the painting for the painting to remain in equilibrium? Make sure your expression is simplified (reduced) and tangible. 9. Two physics students, Brad and Janna, are carrying a 60 kg uniform metal tube 4.5 m long on their shoulders. Jana's shoulder is positioned 1.5 m from one edge, while Brad's shoulder is positioned at the opposite edge. Find the forces that the tube exerts on each of their shoulders assuming that they are of the same height. 10. A uniform plank ($m = 4.0$ kg and $l = 4.0$ m) is in equilibrium with a block of mass M = 20 kg suspended 1.0 m from the left tip. Find the force that each support exerts on the plank.

Transcript

00:01 In this problem, you have a uniform beam attached by a cable that attaches to the wall point b, and it's also attached to point a to the wall.

00:12 Point a is going to be our pivot point.

00:15 We're going to be calculated our torques around that.

00:19 The first thing it wants, and it does give the angles in different places, and we are told that we can use the sum, call that psi, the sum.

00:31 We'll need that later on, but one step at a time.

00:33 Time.

00:34 First thing is the free bi diagram, which i've noticed that you have chosen correctly.

00:39 Tension pulls along the cable.

00:45 Technically this mass, it's pulling, it gives a generator tension in this cable here, which is an equilibrium is equal to the weight.

00:54 So we just write the weight of the hanging mass here.

01:00 And we have the weight of the beam itself, which should be in the center, which actually i think i'm off by a little bit here in terms of that.

01:12 But that's a uniform object.

01:14 It's going to be in the center of the object.

01:19 F sub b, that'll be the weight of the beam, f subab, the weight of the mass, hanging mass, t's the tension, sy and sx are the reaction forces at point a.

01:31 That we'll be looking for all, we'll be looking at those two, and for the tension in the and we'll just do a traditional xy axes.

01:40 Nothing fancy.

01:43 Okay.

01:45 Now, part b wants the lever arm, or another term moment arm for the weight of the beam, this force here.

01:59 Now, because i got other things in here, let me redraw it just for this one.

02:02 I'll redraw it for readability's sake.

02:10 There's fp.

02:12 I'll put in the, i'll at least keep the other ones, but i'm going to leave s, y, and s, x off just for, because i need to draw in here.

02:25 Okay.

02:27 The moment arm, the lever arm.

02:30 What is that? that's the perpendicular distance from the line of action of the force, which is just an infinite line to the force.

02:40 So it's a perpendicular distance from the line of action of the force to rotation axis.

02:44 So that means it makes a 90 degree angle.

02:47 So here, here's my 9 degree angle.

02:51 This is our perpendicular for fb.

02:55 This is the lever arm, moment arm, whatever you're used to having a call.

03:07 And now, geometrically, to get it, this, remember, this angle in here is theta.

03:16 This hypotenuse is l over 2, half.

03:19 And don't, just because i've drawn it this way, don't worry about spacing or anything like that that should be more.

03:27 We're not worrying about that's just a matter of drawing.

03:35 Everything is fine.

03:37 You know, if you do write down the center here and so everything's fine.

03:42 So our perpendicular for fb is going to be, if you do cosine theta, that's going to be our perpendicular over l over 2.

03:51 So that's going to be l over 2, cosine theta.

03:56 And i'm just going to write off to the side the sense of the rotation, the torque, the sign of the torque.

04:02 What way would this force want to turn this object if acting alone? and he would want to turn this clockwise.

04:09 So that's a negative torque.

04:11 So i'll need that later on, but it's not part of this.

04:14 This is the answer they wanted this part, but that's something for future.

04:19 So that was part b.

04:23 Part c, part c, and actually, i can, yeah, i probably can do it all on one diagram now.

04:31 Well, actually, let me not.

04:33 Let me, it'll get too busy in the bottom.

04:36 So let me draw this again.

04:42 Keep all the forces just for completeness sake, except for the s, the s, x, and s, y, because they get in my way for what i'm trying to do.

04:53 Visually, i mean, that's what they're.

04:55 Okay.

04:57 So those are those forces, again, leaving off the other two, just for completeness sake.

05:03 This length is l.

05:05 Now we want the moment arm for fm.

05:13 So there is the line of action.

05:15 And that means that this, this is the, i can put arrows on this.

05:23 This is the moment arm for fm.

05:29 And again, this angle is theta.

05:32 So same thing, coscent theta is the moment arm over the hyponnas, which is now l.

05:38 So our perpendicular fm is l cosine theta.

05:44 And again, getting a sense of the, getting the sign of the torque that we're going to need, what way would it turn it? in fact, you know, also clockwise.

06:03 So that one is negative also.

06:05 Again, that's not part of the answer, but i'm just getting ready for what's coming next.

06:12 All right, part d.

06:15 Now it wants the formula for the tension.

06:18 But before we do that, we need to know it's moment arm.

06:22 And i hope i gave myself enough room.

06:26 Let me get some of this out of the way here.

06:29 Okay.

06:31 All right.

06:32 Here is the line of action for the tension force.

06:39 Here is the moment arm.

06:41 Notice a 90 degree angle.

06:42 Also, shortest distance on the rotation axis to a line of action.

06:47 Anything else is longer than that.

06:49 So however you want to look at it.

06:50 Perpenicular distance, shortest distance, same thing.

06:53 Okay, so this is our perpendicular for t.

06:58 And we know this, we can see the right triangle, so we know that hyponus is l.

07:01 We've got to have an angle.

07:02 Now, let's not forget.

07:05 This angle is phi.

07:07 I have it, i did put it, yeah, i got it there.

07:10 I should probably have kept this in here just for completeness...

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