Two blocks A and B are connected by a slender rod of...

Two blocks A and B are connected by a slender rod of negligible weight as shown in Figure 6. The coefficient of static friction is 0.3 between all surfaces of contact, and the rod forms an angle ? = 30°. a) Show that the system is in equilibrium when P = 0 b) Determine the largest value of P for which equilibrium is maintained. W = 10N W = 10N

Transcript

00:01 Okay, so for this, we need to show that when this downward force p is zero, that this is in equilibrium.

00:11 So the way that i'm going to do that is by figuring out the maximum value for p, and then showing that zero is less than that maximum value.

00:22 So once p overcomes that maximum value, it'll start to move.

00:27 So to do this, first i'm going to evaluate this mass up here.

00:31 These two masses are attached by a rigid rod that it can rotate.

00:37 And so the free body diagram for mass b looks a little something like this.

00:43 We've got its weight force down.

00:45 We've got a friction force to the right, and we've got the normal force up, and then there's this tension force from the bar that would normally be pulling it forward.

00:56 When this is in equilibrium, all these forces cancel out.

01:00 So looking at the x direction, we're going to have a force to the right, that is the friction force.

01:07 And we have the force to the left, which is the x component of this fab, the tension force in that rod.

01:16 And these are going to be equal to zero in the x direction.

01:19 So i can plug in for a friction force.

01:24 This is going to be mu times the normal force, whatever that happens to be, which we'll find it a bit.

01:28 And then i'm going to solve this for the tension force here, fab, in terms of this other stuff.

01:39 I know mu and sine of 30 degrees is 1⁄2, so mu is 0 .3, so this becomes 0 .6 or 0 .6 n is the normal force.

01:57 In the y direction, we are going to have three forces.

02:00 We've got the normal force up.

02:02 We've got the weight force down, and then we've got the y component of this tension force, which i can then plug in this value in there to then have one equation with one unknown.

02:16 So i know the weight force is 10 newtons, and i can plug in for everything else.

02:21 So then solve this equation for the normal force.

02:26 I get normal force factored out of this, so this becomes one minus.

02:34 0 .6 times cosine 30 or 0 .52 quantity on this side.

02:40 This 10 newtons goes on the other side, divide both sides by this...

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