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[SOLVED] Tractor up a slope |
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| Nov19-03, 06:51 PM | #1 |
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[SOLVED] Tractor up a slope
The question is as follow...
A farm tractor tows a 4300kgs trailer up an incline 26 degrees aboe the horizontal at a steady speed of 3.0m/s. What force does the the tractor exert on the trailer? I've read the section over in which this problem is included in... the section is over find the accerlation and force of in 2 dimensions. well if i'm not mistaken...the accerlation of this problem is 0. F=ma F=4300(0)=0? Doesn't make sense OR if the tractor is pulling this tailer...the tractor is obviously exerting the force of the trailer and then some if it's moving it by 3m/s Am I going in the rigth direction? |
| Nov19-03, 07:08 PM | #2 |
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Is the acceleration (of the trailer) = 0? YES!
SO... does F = MA = 0? YES! But you need to understand what F=MA means. It really should be written as: Fnet = MA The net force is zero. But there are several forces acting on the trailer. What are they? |
| Nov19-03, 07:09 PM | #3 |
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Fy and Fx
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| Nov19-03, 07:25 PM | #4 |
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[SOLVED] Tractor up a slope |
| Nov19-03, 07:25 PM | #5 |
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In the Y direction, the total force is 0.
But in the X Direction, the Force is 18472.960125691722371458233543339 N. |
| Nov19-03, 08:17 PM | #6 |
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gravity is pulling it down |
| Nov19-03, 08:33 PM | #7 |
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So... what is the component of the weight along the incline? |
| Nov19-03, 08:35 PM | #8 |
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-FySin(26) + FxCos(26)?
I don't know |
| Nov19-03, 08:46 PM | #9 |
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The weight of the trailer is (mass)x (g)= mg.
The component of the weight along the incline is mg sin(26) pointing down. (Draw a picture!) That must be balanced by the force of the tractor pulling the trailer up the incline. So... Forcetractor pulling trailer = mg sin(26) |
| Nov19-03, 09:36 PM | #10 |
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18472.96n
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| Nov19-03, 11:35 PM | #11 |
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