[Questions 9-10] A horizontal uniform bar of mass 260 kg and...

[Questions 9-10] A horizontal uniform bar of mass 260 kg and length L = 3.0 m is placed on two supports, labeled A and B, located as shown in the diagram. A block of mass 40 kg is placed on the right end of the bar. 9. Find the normal forces FA and FB exerted on the bar by the supports. Use g = 9.8 m/s². [Answer: FA = 588 N; FB = 2352 N] 10. If the 40 kg block is to be replaced with another mass, what is the minimum mass m that would cause the bar to tip over? [answer: m = 130 kg]

Transcript

00:01 In this video, i'm going to be looking at a system in static equilibrium.

00:04 Okay, so what we have is a beam, a uniform beam.

00:09 It's being supported at two points by a support a and support b.

00:17 Okay.

00:18 This total beam has a length l of 3 .0 meters.

00:25 Okay, so that's from end to end.

00:28 That's l.

00:30 Support b is a distance of db equals 1 .0 meters from the right end of the beam.

00:42 Okay, support a is located at a distance of 0 .5 meters from the left end of the beam.

00:52 The distance between supports a and b here is, i'll just call that d -a -b.

01:02 D -a -b equals 1 .5 meters.

01:09 Okay, so we can see that 1 .5 plus 1 .45 equals l, our total length of the beam.

01:17 This beam has a little box on the end of it with some mass m equals 40 kilograms.

01:24 Okay, and the mass of the beam itself, m is 260 kilograms.

01:32 Okay, and what i want to find are.

01:34 The forces that support a and support b exert on this beam to keep it in its static equilibrium state.

01:42 Okay, and for this to be true, the sum of all the forces acting on the beam must equal zero.

01:49 Okay, and the sum of all the torques acting on the beam must equal zero.

01:55 Okay, so let's look at our forces first.

01:57 We have two forces acting in the positive direction.

02:00 That's going to be the force from beam a, and the force from beam b.

02:06 Okay, and in the negative direction, we have the mass of the beam times the acceleration due to gravity, minus the max of the box times the acceleration due to gravity.

02:19 Okay, we're going to do the same thing for torque.

02:21 And before we start talking about torque, we're talking about rotation.

02:24 We need to choose a pivot point.

02:26 I'm going to choose it to be at the center of a mass of the beam.

02:30 It's a uniform beam, so that's right in the center.

02:32 Okay, now that we're at the center, that gravitational force due to the beam itself will not factor in because it's right at the pivot point.

02:42 Okay, so we're just going to have to consider three forces.

02:45 That's going to be f -a, f -b, and the gravitational force due to the box.

02:49 Okay, and when we're talking about torques, we're talking about rotation, so we need to choose an orientation.

02:55 I'm going to call forces that tend to cause a counterclockwise rotation.

02:59 I'm going to call that the positive direction.

03:01 And forces that tend to cause rotation in the clockwise direction.

03:06 That's going to be my negative direction.

03:08 Okay, so let's look at our forces that will cause a clockwise rotation.

03:12 They only have one, so only one force contributing to positive torque.

03:19 So that's going to be the force from support b times support b's distance from the pivot point.

03:26 Okay, where our pivot point is at 1 .5 meters.

03:32 Is one meter from the end of the beam, so the distance to the pivot point is 0 .5 meters.

03:39 All right.

03:39 Now let's look at my forces that cause a negative rotation.

03:45 The rotation in the clockwise direction, that's going to be force of a.

03:49 Okay, so minus fa, okay, times its distance to the pivot point.

03:56 So that's going to be one meter...

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