2 this problem refers to figure 2 a 500 lb crate b is...

2) This problem refers to figure 2. A 500-lb crate B is suspended from a cable attached to a 40-lb trolley A which rides on an inclined I-beam as shown. At the instant shown, the trolley has an acceleration of 4.2 ft/s² up and to the right. At the instant shown, the cable attached to the trolley A is vertical. Determine the acceleration of B relative to A. $a_{B/A} = $ ft/s² up / down / left / right (circle one) Determine the tension in the cable CD. $T_{CD} = $ lb

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00:01 So in this problem we have a 500 pound crate that is suspended from the cable attached to a 40 pound trolley, which rides on an inclined eye beam.

00:12 So we want to find the acceleration of b relative to a and also the tension in cable c, d.

00:20 So we're given that the weight of b is 500 pounds.

00:27 The weight of a is 40 pounds.

00:30 And the acceleration is given as 1 .2 feet per second squared.

00:40 So first let's define a coordinate system and draw the free body diagram for b.

00:50 So we have its force of gravity down, it's just its weight, just the tension up.

01:02 So first let's find the acceleration of b relative to a.

01:13 So the acceleration of b is equal to acceleration of a plus the acceleration of b relative to a.

01:26 And using newton's second law, which is just the sum of the forces in the x is equal to m .a .x.

01:38 So, okay, and this is coming from the diagram where we have the angle between the beam and the horizon is 25 degrees.

02:05 And because in a we have a situation where the a of a is at that, and then the acceleration of b over a is like that.

02:25 So that's where this cosine of 25 degrees is coming from.

02:30 And then we want to solve for the acceleration of b over a.

02:34 So we're just going to bring this over and the masses cancel because they're the same here...

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