(II) A 265-kg load is lifted 18.0 m vertically with an acceler...

Question

(II) A 265-kg load is lifted 18.0 m vertically with an acceleration a = 0.140 g by a single cable. Determine

(a) the tension in the cable;

Accepted Answer

Hey, everyone. Welcome back in this problem. A construction crane lifts a steel girder with a mass of 800 kg at an acceleration of 0.2 G to a certain height on a building site. We're asked to calculate the tension in the lifting cable while the girder is accelerated. We have four answer choices. All of Newton's option A 1.6 multiplied by 10 to the exponent three. Option B 7.8 multiplied by 10 to the exponent three. Option C 8.0 multiplied by 10 to the exponent three and option D 9.4 multiplied by 10 to the exponent. So we're looking for the tension. OK? That's a force. So let's go ahead and draw free body divers and identify all the forces acting on this girl. We have our gutter and we know that it's going to have a force of gravity acting downwards. It's one of those forces that we typically always have and then the construction crane is lifting it. So the tension force is going to be acting straight upwards right now, the crane is lifting it. So we're gonna choose upwards to be our positive direction and we know that we have an acceleration in that direction of 0.2 G, right? So we have some information about forces, we have some information about acceleration. Let's go ahead and use Newton's second law which tells us that the sum of the forces is going to be equal to the mass multiplied by the acceleration. In this case, we're talking about the vertical forces and the vertical acceleration. That's where all of our motion is. We've chosen up to be our positive direction. And so the sum of the forces is gonna be that tension force that's acting upwards minus the force of gravity that's acting downwards. And this is equal to our mass multiplied by our acceleration. We can rearrange uh our tension force T is gonna be the mass multiplied by the acceleration plus the force of gravity. I recall that the force of gravity is equal to M multiplied by G. So we have T is equal to MA plus MG. OK. So we have the masses. We have some information about the acceleration. We know the acceleration due to gravity G. So let's start by filling in what we know about the acceleration. OK. So we have our attention force. This is gonna be equal to the mass M. And our acceleration we're told is 0.2 G. So we have M multiplied by 0.2 G plus M multiplied by G. We can factor M and G from both terms on the right hand side, we have MG multiplied by 0.2 plus one. And so this simplifies down to just 1.2 multiplied by the mass multiplied by the acceleration due to gravity. And that is our tension force. T remember that is what we're trying to calculate. So now that we've simplified it, we're gonna go ahead and substitute in our values. OK? We could have substituted in our values a few lines earlier before we rearranged and simplified, we would get the same answer. But this is nice. We've simplified it down and now it's more straightforward to substitute in those values. We have 1.2 multiplied by the mass, which we're told is 800 kg multiplied by the acceleration due to gravity 9.8 m per second squared. And when we work all of this though, we get attention t of approximately 9000 400 sorry, no period there. The tension is going to be equal to 9408 newtons, which is approximately equal to 9.4 multiplied by 10 to the exponent three newtons. If we put it into scientific notation with two significant digits, like the answer choices have. OK. So that is the tension in the lifting table while the girder is accelerating. If we compare this to what we had in the answer choices, we can see that the correct answer is going to be option D. Thanks everyone for watching, I hope this video see you in the next one.

Struggling with Physics?

(II) A 265-kg load is lifted 18.0 m vertically with an acceleration a = 0.140 g by a single cable. Determine
(a) the tension in the cable;

Watch next

Vertical Forces & Acceleration practice set

Struggling with Physics?

(II) A 265-kg load is lifted 18.0 m vertically with an acceleration a = 0.140 g by a single cable. Determine(a) the tension in the cable;

Watch next

Vertical Forces & Acceleration practice set

(II) A 265-kg load is lifted 18.0 m vertically with an acceleration a = 0.140 g by a single cable. Determine
(a) the tension in the cable;