How Much Tension Is Needed to Lift a 1800 kg Car Upward at 0.60 m/s²?

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SUMMARY

The discussion centers on calculating the tension required in a rope to lift an 1800 kg car upward with an acceleration of 0.60 m/s². Participants emphasize using a free body diagram to analyze the forces acting on the car, specifically the tension in the rope and gravitational force. The correct formula derived is T - mg = ma, where T is tension, m is mass, and g is the acceleration due to gravity. The final calculation confirms that the tension must account for both the weight of the car and the additional force needed for upward acceleration.

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This discussion is beneficial for physics students, educators, and anyone interested in understanding the principles of force and motion, particularly in the context of tension in ropes and lifting objects.

confusedaboutphysics
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How much tension must a rope withstand if it is used to accelerate a 1800 kg car vertically upward at 0.60 m/s2?

could someone tell me how to start this problem? thanks so much!
 
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Consider a free body diagram around the car. Net force = mass * accelleration. So you have the tension in the rope and gravity acting on the car, and the total of those must give you an upwards accelleration of 0.6m/s^2
 
i'm still confused with this problem..i thought i could just multiply 1800kg by .60 m/s^2? but i don't get the right answer...

help please!
 
confusedaboutphysics said:
How much tension must a rope withstand if it is used to accelerate a 1800 kg car vertically upward at 0.60 m/s2?

could someone tell me how to start this problem? thanks so much!

Is this problem neglecting air friction and mass of the rope? Also, what's pulling that rope?
 
Last edited:
it doesn't say...its a webassign problem
 
confusedaboutphysics said:
it doesn't say...its a webassign problem

Ok, then what is the answer, if given? Because I got what you got too. I might be missing something
 
sorry...i don't have the answer either..it's due tomorrow night..hmmm...i wander why it won't work...
 
confusedaboutphysics said:
i'm still confused with this problem..i thought i could just multiply 1800kg by .60 m/s^2? but i don't get the right answer...

help please!

You don't get the right answer even though you don't have the right answer? lol
 
webassign is a homework program on the internet...it tells you right away if you're wrong or right...but doesn't give you the right answer until after its due or if you get the answer right.
 
  • #10
I think what Kazza is saying is \vec{F_T} - mg = m\vec{a}. If that's correct then you can think of a free body diagram of the rope with the upward force equal to ma, and the downward force equal to mg. But don't take my word on it :biggrin:
 
Last edited:
  • #11
thanks cscott! its correct!
 
  • #12
Woo! I give 94.8% credit to Kazza :wink:
 

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