An elevator cab that weighs 25.2 kN moves upward. What is the tension in the cab

Question

An elevator cab that weighs 25.2 kN moves upward. What is the tension in the cable if the cab's speed is (a) increasing at a rate of 1.21 m/s2 and (b) decreasing at a rate of 1.21 m/s2?

Explanation / Answer

The majority of this answer is to help you to build a conceptual basis of why the answer is correct, if you really want to understand this question, I would suggest reading through the entire thing, however, IF YOU SIMPLY WANT THE ANSWERS (which I totally understand), I have added a re-cap at the end which has a greatly simplified process. I think that the best way to go about this is to look at the problem in equilibrium first, then add (or subtract the accelerations). The tension in the cable is measured as a force (in Newtons), so the tension in the cable when the cab is at rest, or traveling at a constant velocity would be equal to the weight of the cab (25.2kN). We know that F=m*a, and we are given the accelerations, so we need to find the mass. We also know that weight (W) is mass*gravity, where the weight of the car is 25.2kN and we will assume that gravity is 9.81 m/s^2 (standard acceleration of gravity near the surface of the earth). Solving for the mass of the cab, we get 2570 kg. To find the tension in the cable, we must first find the force due to the acceleration (or deceleration) of the car then add (or subtract) it from the tension due to gravity. Lets first look at (a) with an increasing velocity of 1.21m/s^2. Using F=ma, the force that we get F=(2570kg)*1.21m/s^s==> F=3.12kN. Now lets think about whether we should add or subtract this from the tension due to gravity. If you imagine riding in an elevator upwards, when it first starts moving, it feels like you're heavier, like there's something pushing down on you. This feeling suggests that when the elevator is accelerating (as in case a) the force is added to the tension due to gravity, so the total tension in the cable would be Tt=Tg+Ta where Tt is the total tension, Tg is the tension due to gravity, and Ta is the tension due to the acceleration of the cab. Tt=25.2kN+3.12kN= 28.32kN For case b, if you imagine reaching the floor that you want to in the elevator after traveling upward, it almost feels like you're floating. You might imagine that you would subtract this force from the tension due to gravity, and you would be right in your assumption. Because the accelerations acting on the same mass are equal and opposite, it is true that the forces generated are also equal and opposite. Thus, Tt=Tg-Ta==> Tt=25.2kN-3.12kN=22.08kN Re-Cap Give: Weight=25.2kN, The elevator is moving upwards. Relevant equations: F=m*a, W=m*9.81 Find: the tension in the cable a)when the cab is accelerating at 1.21m/s^2 b)decelerating at 1.21m/s^2 Solution: m=W/g= 25.2kN/9.81m/s^2= 2570kg a)Tt=W+m*a=25200N+(2570kg*1.21m/s^s)=28.32kN

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