As shown in Fig. 4–48, five balls (masses 2.00, 2.05, 2.10, 2....

Question

As shown in Fig. 4–48, five balls (masses 2.00, 2.05, 2.10, 2.15, 2.20 kg) hang from a crossbar. Each mass is supported by '5-lb test' fishing line which will break when its tension force exceeds 22.2 N (5.00lb). When this device is placed in an elevator, which accelerates upward, only the lines attached to the 2.05 and 2.00 kg masses do not break. Within what range is the elevator's acceleration?

Five hanging balls with varying masses are supported by fishing lines, illustrating tension forces in an accelerating elevator.

Accepted Answer

Welcome back everyone in this problem in a classroom demonstration, five different masses 1 kg, 1.21 0.41 0.6 and 1.8 kg are suspended from a horizontal bar using strings as shown in the figure below. Each string has a maximum tension of 25 newtons before breaking. When this setup is placed in a lift accelerating upward, only the strings attached to the 1 kg 1.2 kg and 1.4 kg masses remain intact. Determine the range of acceleration for the elevator. For our answer choices. A says the acceleration is between 13.9 and 15.6 m per second. Squared B says it is between 5.83 and 8.06 m per second squared C between 5.83 and 15.6 m per second squared and D between 8.06 and 13.9 meters per second squared. Now let's first ensure we understand what we're looking for here. We want to find a range of acceleration for the elevator before we find a range of acceleration. Let's try to understand what's happening to the masses when they're accelerating upward. So I'm going to draw a free body diagram here to the right. OK. And when our masses are accelerating upward, the tension of the string, let's call it ft is acting upwards on the mass while the mass is on weight MG is acting downwards on the mass. Now, when we think about forces, what does it have to do with acceleration? Well, we can recall that by Newton's second law, the force is directly proportional to acceleration by the formula that says the sum of forces equals the mass multiplied by the acceleration. Now, if we are going to find a range of acceleration range of the acceleration and compare accelerations, we will need a formula to figure out what it is. So let's resolve our forces in the Y direction and apply it to this formula to see if we can solve for our acceleration. A. Now as we said vertically, if we take the opper direction to be positive, then the force is acting upwards KFT minus the first acting downwards MG is going to be equal to M multiplied by A. That means the software acceleration A, we can divide through by M. So A is going to be equal to attention FT divided by M minus or weight MG divided by M where M cancels M and that leaves us with G. So this is a formula that we can use to figure out our acceleration. Now, what do we know about acceleration. Well, since the string broke for the 1.6 kg mass, we know that the required tension to accelerate that mass was more than 25 newtons. Likewise, since the string didn't break for the 1.4 kg mass, we know that the required tension to accelerate that mass was less than 25 newtons. OK. So in essence, then we know that since the force, OK, the force for our 1.6 kg mass was more than 25 newtons. OK? Then when we think about acceleration, that means our acceleration would also have to be greater since force is directly proportional to acceleration. Another way we can think about it is that when we're talking about acceleration, a heavier mass will decrease our acceleration. OK. So what that tells us then is that four or 1.6 kg mass, its acceleration is going to be less than the 1.4 kg mass. Keeping in keeping in mind the same tension in the rope. So since our acceleration, OK, for our 1.6 kg mass is going to be less than our acceleration for 1.4 kg mass, then our acceleration is going to be between the acceleration for 1.4 kg mass and the 1.6 kg mass. And as we said, that makes sense because the string breaks at 1.6 kg, but it's still intact for the 1.4 kg weight. So if we're going to figure out our our range for acceleration. Basically, we can find acceleration at 1.6 kg. Let me put that in red and then find acceleration for 1.4 kg and that will give us our range. So let's start with our 1.6 kg weight for 1.6 kg. Based on our formula then or acceleration is going to be equal to or maximum tension 25 newtons divided by the mass which is 1.6 kg minus acceleration due to gravity which we can take as 9.8 m per second squared. Now we can put this expression into our calculator. OK? And when we do that, we find that for our 11.6 kg mass, the acceleration is going to be 5.825 m per second squared, which rounded to the nearest 100 is 5.83 m per second squared. Next, we can do the same thing for our 1.4 kg. So 11.4 kg, the acceleration is going to be equal to 25 newtons divided by 1.4 kg minus 9.8 m per second squared. Again, if we put this expression into a calculator, we should get the acceleration to be equal to 8.057 m per second squared, which rounded to the nearest 100 would be 8.06 m per second squared. What does this tell? Us. Well, no, this tells us then that our acceleration is going to be. Let me put that in back here. Our acceleration is going to be between 5.83 m per second squared and 8.06 m per second squared. Therefore, B is the correct answer. Thanks for watching everyone. I hope this video helped.

Key Concepts

Tension in a Rope

Newton's Second Law of Motion

Weight and Gravitational Force

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