The tractor is used to lift the 150 kg load b with the 24 m...

The tractor is used to lift the 150-kg load B with the 24-m-long rope, boom, and pulley system. The tractor travels to the right at a constant speed of 5 m/s. When $s_A = 0$, $s_B = 0$. (Figure 1) Determine the tension in the rope when $s_A = 5$ m. Express your answer to three significant figures and include the appropriate units. $T = \text{Value } N$

Transcript

00:01 Hi everyone, so what we have here is a tractor that is moving at a velocity to the right of 4 meters per second.

00:08 And attached to that tractor is a rope which goes over a pulley and is being used to lift this mass b.

00:15 We know that the mass b has a mass equal to 150 kg, and this mass is being lifted over with the pulley.

00:29 What we're asked to solve for is to find the tension when this distance s -a, which is defined as the distance from the mass to the tractor, is equal to five meters.

00:39 So what we're going to be doing to solve this problem is defining the length of the rope in terms of s -a and s -b, and then applying newton's second law to solve for the acceleration of the block to solve for the tension, s .a.

00:58 Is equal to 5 meters.

01:01 So let's go ahead and draw a diagram of just this rope.

01:06 So this rope shown here, which is going to be equal to a length equal to 24.

01:14 We know that this distance from the box to the tractor is s .a.

01:21 And we know the distance from the ground to the box is equal to sb.

01:26 Therefore, this height will be equal to a value, but we know that the total distance from the top of the pulley down to the ground is equal to 12 meters.

01:38 Therefore, this distance here is just going to be equal to 12 minus sb.

01:44 What we can do is we can say that l, which is equal to 24, will be equal to 12 minus sb.

01:56 Plus the square root of s -a -squared plus 12 squared.

02:01 And that is just the hypotenuse of this triangle.

02:06 Next, what we can do is go ahead and differentiate this entire expression with respect to time to obtain velocity.

02:14 So by taking the derivative of this expression here with respect to time, we see that dl, d -t, in terms of s -a and s -b, is going to be equal to zero...

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