A 500 kg air conditioner sits on the flat roof of a building. The coefficient of static friction between the roof and the air conditioner is 0.90. A massless rope attached to the air conditioner passes over a massless, frictionless pulley at the edge of the roof. In an effort to drag the air conditioner to the edge of the roof, four 100 kg students hang from the free end of the rope, but the air conditioner refuses to budge. What is the magnitude of the rope tension at the point where it is attached to the air conditioner?

Question

Accepted Answer

Hey, everyone. So this is a pretty straightforward Newton Second law problem. Let's see what they're asking us. We have an 80 kg crate lying on a level bench. The static friction coefficient between the crate and the bench is 800.85 A cable with negligible mass is tied to a crate and then directed through a massless and frictionless pulley at the edge of the bench in an attempt to drag the crate on the bench. 3 20 kg masses are tied to the other end of the cable. The crate is not dislodged. We are asked to determine the the cables tension at the point where it is tied to the crate. Our multiple choice answers here are a newtons. B 784 newtons C newtons or D 1370 newtons. So the first thing we're going to do is draw out what is happening in this problem and then identify the forces. So we have a crate, it's 80 kg, it weighs 80 kg. We have a cable that is attached to the crate, goes around a pulley and then we have three weights attached to the bottom of that end of the pool that are each 20 kg. So when we look at the forces acting in this system, we've got the friction force negative to the um direction of, of motion or of attempted motion. In this case, we have our tension force on the cable. We have the tension force acting this way on the cable and then acting sorry in the um neg in the positive Y direction on the cable there. And then we have our weight acting on the 320 kg Masses. And then we also have weight of that crate and the normal force. So we have a, a lot going on. But if we can recall from our um tension how tension works in cables, we know that the tension in this section of the cable is in a section of the cable that is um after the pulley, after the bench is equal to the tension that is between the pulley and the crate. So we can take the 320 kg masses as a um single particle and use Newton's second law for forces in the Y direction to solve for that tension. So we can recall Newton's second lot is the sum of the forces is equal to mass multiplied by acceleration. The some of the forces in the Y direction are tension and then minus the weight because the weight is acting in the negative direction. And we know that the crate is not dislodged, there is no movement. Therefore, there is no acceleration. So that side of the equation is equal to zero. So now we have our tension is equal to our weight. And we can recall that our weight is given as mass multiplied by gravity where the mass is three multiplied by 20 kg then multiplied by gravity which is our constant 9.8 m per second squared. Plug that into our equation or into our calculator and get 588 newtons. And so that is the answer to this problem that aligns with answer choice. A it's all we have for this one and we'll see you in the next video.

Verified Solution

Key Concepts

Static Friction

Tension in a Rope

Newton's Second Law