How Do You Calculate Tension and Acceleration in a Towing Scenario?

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In a towing scenario, a breakdown van with a mass of 2000 kg tows a 1200 kg car, facing resistances of 800 N and 240 N, respectively, and a driving force of 2320 N. The acceleration of both vehicles is calculated to be 0.4 m/s², with the tension in the tow bar at 720 N. When moving up an incline with a sine angle of 1/20, the acceleration changes, and the correct approach involves resolving forces along and perpendicular to the incline. The final acceleration is determined to be 0.09 m/s², indicating a decrease in speed as they ascend. A recommended strategy for similar problems includes drawing a diagram and resolving forces accordingly.
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Mechanics Question. IV :)

A breakdown van of mass 2000 kg is towing a car of mass 1200 kg along a straight horizontal road. The two vehicles are joined by a tow bar which remains parallel to the road. The van and the car experience constant resistances to motion of magnitudes 800 N and 240 N respectively. There is a constant driving force acting on the van of 2320 N. Find

(a) the magnitude of the acceleration of the van and the car,

(b) the tension in the tow bar.

The two vehicles come to a hill inclined at an angle α to the horizontal, where sin αlpha= 1/20 . The driving force and the resistances to the motion are unchanged.

(c) Find the magnitude of the acceleration of the van and the car as they move up the hill and state whether their speed increases or decreases.

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Hey...
I get the first 2 questions... with the answers of 0.4m/s and 720.

however (c) .. i don't really get... i would think of using the sin 1/20 in my answer [along with F=ma where m is 3200 (sum of the two cars' masses) ]... but the model answer contradicts this... they don't use sin or cos...
could someone please help me out on this one... :)

Thanks a lot.
 
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Consider just a single mass m on a hill (inclined plane) inclinded at an angle \alpha. Can you resolve the forces of weight and friction in two mutually perpendicular directions: along and perpendicular to the incline?
 
that is exactly what i was thinking... solve the force horizontally and then vertically... however; that does not give me the answer that is in the mark scheme... 0.09 :confused:
 
turnstile said:
that is exactly what i was thinking... solve the force horizontally and then vertically... however; that does not give me the answer that is in the mark scheme... 0.09 :confused:

How have you resolved? Can you show your working please.

(The answer in the mark scheme is correct)

~H
 
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ah.. i got it now..
the way I've done it now is;

3200a= 2320-1040-3200gx 1/20

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Is there any golden rule of thumb you can suggest me when working with these sort of questions? :)
 
turnstile said:
3200a= 2320-1040-3200gx 1/20

Yep, that's right. :smile:

Question said:
state whether their speed increases or decreases

Don't forget to answer this bit!

turnstile said:
Is there any golden rule of thumb you can suggest me when working with these sort of questions? :)

The first thing I always do is draw a diagram of the incline with all the forces on. The I resolve all the forces so that they are either perpendicular to the incline or parallel to it. After that its just summation of forces.

~H
 
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