i-how-much-tension-must-a-rope-withstand-if-it-is-used-to-accelerate-a-1210-kg-car-horizontally-alon

Name: i how much tension must a rope withstand if it is used to accelerate a 1210 kg car horizontally alon
Uploaded: 2019-05-15
Duration: 39 s
Channel: Numerade
Description: I) How much tension must a rope withstand if it is used to accelerate a 1210-kg car horizontally along a frictionless surface at 1.20 m/s$^2$ ?

Question

I) How much tension must a rope withstand if it is used to accelerate a 1210-kg car horizontally along a frictionless surface at 1.20 m/s$^2$ ?

Averell Hause · Accepted Answer

So we have the mass for the car one thousand, but one thousand two hundred ten kilograms and let me have the acceleration of the car one point two zero meters per second squared. And then we want the force tension needed in order to accelerate this car by by a magnitude of one point two meters per second squared. So this is simply going to be m times, eh? One thousand two hundred and ten times one point two. And we&#x27;re getting a tension of one thousand four hundred and fifty two Nunes and this would be our final answer. That&#x27;s the end of the solution. Thank you for watching.

Eric Mockensturm · Answer

we have were asked how much tension must rope. We&#x27;re standing for jews to accelerate a 960 kg car at a rate of 1.5 1.2 m/s squared. So just just like problem one, we have a the weight of the car, the normal force. And again, we&#x27;re assuming that the friction force or whatever drag force, ruling resistance or whatever it is, um, is zero, so that in the X direction, all we have is the force in the X direction. At times the mass equals the mass times acceleration in the X direction. And so multiplying that out, we get that, that is 1150 newtons. So again, it&#x27;s quite a large force because we have quite a lot of mass here. Now again, this is kind of would be the minimum force. Again, if the forces in the exact same direction as acceleration, it had a component upward than the magnitude would be greater than this, but only the X component would be used to accelerate it. The y component would be used to reduce the normal force

Averell Hause · Answer

the rope and so we can apply these some of forces to the system. We can say that the sum of forces would be equal to forest tension, which would be the only, uh, force on the car. And this would be equal to the mass times go acceleration. So to find forest tension, you simply plug in the mass and the acceleration. So that would be the 1210 kilogram car multiplied by the acceleration of 1.2 meters per second squared. And we find that the maximum force tension that the rope can handle would be 1452 Newton&#x27;s. And so this would be our final answer. That is the end of the solution. Thank you for watching.

Eric Mockensturm · Answer

Yeah, we&#x27;ll ask how much tension muster rope withstand if it&#x27;s used to accelerate a 1200 kg car vertically upwards at 0.8 m per second squared. So we have a free body diagram of our car here, and we have the rope that&#x27;s accelerating it upwards, the acceleration of the car and then the weight of the car, pulling it back down. Um Writing Newton&#x27;s laws for this, we get the tension minus the weight equals the mass times the acceleration of the, of the vehicle. Um the weight is just MG. So moving that to the other side, taking point out the and we get the detention is the mass times the sum of the acceleration plus gravity gravitational acceleration. We know the mass is 1200 kg. Um This value here, we know the acceleration and then we know gravitational acceleration and that means that the tension um that the cable must be able to withstand the tension of about 9420 newtons. So obviously, we want to have a little bit margin of safety there. So we want at least maybe, you know, 1200 newton cable or something. Maybe even more than that, to be careful, um, if there was any kind of, uh, dynamic bloating so that this cable wouldn&#x27;t step, which would be, um, probably not a good for the vehicle.

Darshan Maheshwari · Answer

this question, they&#x27;re asking about how much tension Mr cable with stan. If it is used to accelerate a 1200 kg car vertically upwards at 0.7. Meet the pass again. Square. So, uh, let&#x27;s draw simple figure over here, this is the mosque which we have to accelerate. This is attention. This is the weight of this particular object and it&#x27;s going upwards are 0.7 m per second squared. So the equation of motion is going to look like t minus W is equal to uh, M A. Hence t wait is nothing but MG. So t minus MG is equal to M A. So T is going to be equal to MG plus M A. Uh FM is taken us common. There&#x27;s nothing but I m times G plus. So I&#x27;m just given us 1200 G is nothing but 9.8 and a is given us 0.7. So the acceleration, which is the value of tension which is coming from here is let me grab my calculators or 9.8 plus .7 Times 1200. That&#x27;s 12 612, 12,600 Newtons. is the attention, which the rope has to withstand to have an acceleration of 0.7 m per second squared. Thank you.

Averell Hause · Answer

the force tension so we can draw the car or we could just start as a box going straight down with, of course, BMG and then going straight up would be forced tension. We can say that the sum of forces in the UAE direction equals the mass times acceleration in the wind direction and this would be equal to the force tension minus mg Ah, we&#x27;re going to say that the force tension would then be equal to the mass times gravity plus the acceleration in the wind direction. Uh, this point we can solve. So the mass of the car 1200 kilograms and then multiplied by 9.8 plus 0.7 meters per second squared. And at this point, we can say that the force tension is going to be equal to 1.3 times 10 to the fourth noons and this would be our final answer. That is the end of the solution. Thank you for watching

Averell Hause · Answer

What the force on the rope is going to be the mass of the car times the acceleration Mass is 1210 kilograms, Acceleration is 1.20 meters per second squared. Um, and that&#x27;s gonna be the same as the tension. So just putting that into a calculator, 1210 times 1.2 is 1000, 450 newton&#x27;s. There are only three significant digits, Okay?

Problem

(II) According to a simplified model of a mammali…

Problem 3 Easy Difficulty

I) How much tension must a rope withstand if it is used to accelerate a 1210-kg car horizontally along a frictionless surface at 1.20 m/s$^2$ ?

Answer

1452N

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