ATAR Notes: Forum
VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Physics => Topic started by: Chavi on May 29, 2010, 07:15:59 pm
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Can I please get some help with the following question?
A car of mass 700kg is towing a trailer of mass 500kg, on an incline of 30 degrees to the horizontal.
Friction on the trailer is 400n, friction on the car is 600n. If the engine provides a driving force of 700n, what is the tension in the towbar?
Thanks in advance,
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This appeared at first to be simple but it is abit out of the ordinary here is my attempt at solving it,
I am assuming that the velocity of the car and trailer isnt constant as forces are unbalanced...(more friction than driving force)
Therefore Force going in uphill direction = 700 N
Force (or friction) going downhill = *12000)
=7000N
However These are irrelevant as we are interested in the difference of the net forces acting on the trailer and car
Let up hill be negative and downhill be positive
Net force acting on the car is equal to -700=3400)
Net force acting on the trailer is equal to =2900)
Car has more force than trailer downhill therefore 0 tension in the tow bar as the car and trailer will both be going backwards, i feel as if you are leaving some information out...is the car going downhill or up hill etc...
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must be going downhill, T = 275 N.
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Is it even possible to have MORE friction than the force applied?
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yes going downhill
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Ah, I see.
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is 275N correct? can someone confirm, im getting an answer of 2735N hmmm.
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can anyone confirm at all?
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Can I just ask..is there a formula to find the tension in a string and a circularing object such as one spinning around a pole?
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The tension in a rope spinning something around is the centripetal force.
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Oh sorry I mean like umm...when the ball is banked and the centripetal force is like the horizontal force
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A car of mass 700kg is towing a trailer of mass 500kg, on an incline of 30 degrees to the horizontal.
Friction on the trailer is 400n, friction on the car is 600n. If the engine provides a driving force of 700n, what is the tension in the towbar?
Can someone tell me what answer they get for this one?
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Net force acting on system = 700N + 12000sin(30)N - 1000N
= 5700N down slope
Therefore System accelration = 4.75 m/s^2
Isolate the trailer:
It has an acceleration of 4.75 m/s^2, therefore the Net Force acting on the trailer = 4.75 x 500 = 2375
Therefore, Net Force = Tension + Wx component - Friction
2375 = Tension + 5000sin(30) - 400
Tension = 275N
Can anyone confirm this working out is correct?
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Yes thats what I got as well