Question Video: Calculating Stress in a Rope

Video Transcript

A uniform rope of cross-sectional area 0.600 centimeters squared breaks when the tensile stress in it reaches 7.2 times 10 to the sixth newtons per meter squared. What is the maximum load that can be lifted slowly at a constant speed by the rope? What is the maximum load that can be lifted by the rope with an acceleration of 4.0 meters per second squared?

In this two-part problem, we wanna solve first for the maximum load the rope can lift at a constant speed. We’ll call that 𝐿 sub 𝑠. And then, we wanna solve for the maximum load the rope can lift when that load is accelerating upward at 4.0 meters per second squared. We’ll call this load 𝐿 sub 𝑎. In this scenario, we have a weight we’ve called 𝑊, which at first is being raised at a steady speed by a rope. Knowing the maximum tensile stress the rope can handle before breaking, we want to solve for the maximum load 𝐿 sub 𝑠. If we choose the upward direction to be positive and consider the forces on the weight 𝑊, we can write that the tension force created by the rope is equal to the maximum load, 𝐿 sub 𝑠, that can be lifted at a steady speed. That tension force 𝑇 is equal to the maximum stress the rope can handle multiplied by the cross-sectional area of the rope.

When we substitute in these two values, we’re careful to write our cross-sectional area 𝐴 in units of square meters. That’s to agree with the units of our stress, newtons per meter squared. We see when we multiply these numbers, we’ll get a result in units of newtons. And, to two significant figures, 𝐿 sub 𝑠 is 430 newtons. That’s the maximum load this rope can lift at a steady speed without breaking.

Next, we imagine that our scenario changes a bit in that, instead of our weight being lifted at a constant speed, it’s now accelerating upward at 4.0 meters per second squared. Under these new conditions, we wanna solve for the maximum load that this rope can handle. When we write out the forces on 𝑊 now, we again have the tension force from the rope 𝑇. And subtracted from that is the weight of our load, 𝐿 sub 𝑎. Whereas before, that difference was equal to zero. Now because we have a nonzero acceleration by the second law of motion. It’s equal to 𝑚 times 𝑎, where 𝑚 is the mass of our load, 𝐿 sub 𝑎. Speaking of the mass of this weight, as we think about it, 𝐿 sub 𝑎 is the mass of this weight times 𝑔, the acceleration due to gravity, where 𝑔 we’ll assume is exactly 9.8 meters per second squared.

If we add 𝑚 times 𝑔 to both sides of this equation and we rewrite the tension force in terms of the maximum stress, 𝜎 max, times the area 𝐴, then we now have an expression that says 𝜎 max times 𝐴 is equal to 𝑚 times the quantity 𝑔 plus 𝑎. It might seem like we’re moving backwards because nowhere in this expression do we see a value for 𝐿 sub 𝑎, what we want to solve for. But 𝐿𝑎 is equal to 𝑚 times 𝑔. And we know 𝑔. So if we can solve for 𝑚, then we’ll have solved effectively for 𝐿𝑎.

So back to our force balance equation, rearranging this equation for the mass 𝑚, we know 𝜎 sub max, 𝐴, 𝑔, and the acceleration 𝑎. Then when we plug these values in, again expression our cross-sectional area 𝐴 in units of meters squared, and calculate this term, we find it’s roughly equal to 31.3 kilograms. Going back to our equation for 𝐿𝑎, when we plug this value in for 𝑚 and then 9.8 meters per second squared for 𝑔, our result, to two significant figures, is 310 newtons. That’s the maximum weight that this rope could sustain while accelerating upward at 4.0 meters per second squared.