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Tension problem

by confusedaboutphysics
Tags: tension
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confusedaboutphysics
#1
Oct4-05, 11:16 PM
P: 36
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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Kazza_765
#2
Oct4-05, 11:25 PM
P: 164
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
confusedaboutphysics
#3
Oct5-05, 05:18 PM
P: 36
i'm still confused with this problem..i thought i could just multiply 1800kg by .60 m/s^2? but i dont get the right answer...

help please!!

PhysicsinCalifornia
#4
Oct5-05, 05:44 PM
P: 58
Tension problem

Quote Quote by confusedaboutphysics
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?
confusedaboutphysics
#5
Oct5-05, 05:48 PM
P: 36
it doesn't say...its a webassign problem
PhysicsinCalifornia
#6
Oct5-05, 05:52 PM
P: 58
Quote Quote by confusedaboutphysics
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
confusedaboutphysics
#7
Oct5-05, 06:03 PM
P: 36
sorry...i dont have the answer either..it's due tomorrow night..hmmm...i wander why it won't work...
PhysicsinCalifornia
#8
Oct5-05, 06:12 PM
P: 58
Quote Quote by confusedaboutphysics
i'm still confused with this problem..i thought i could just multiply 1800kg by .60 m/s^2? but i dont get the right answer...

help please!!
You don't get the right answer even though you don't have the right answer? lol
confusedaboutphysics
#9
Oct5-05, 06:19 PM
P: 36
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.
cscott
#10
Oct5-05, 06:21 PM
P: 786
I think what Kazza is saying is [itex]\vec{F_T} - mg = m\vec{a}[/itex]. 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
confusedaboutphysics
#11
Oct5-05, 06:28 PM
P: 36
thanks cscott!!! its correct!!
cscott
#12
Oct5-05, 06:33 PM
P: 786
Woo! I give 94.8% credit to Kazza


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