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Math Help - Mechanics - Dynamics

  1. #1
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    Angry Mechanics - Dynamics

    Is this the right way to work out these answers?

    1 - Car has weight of 1500kg and is travelling along a straight, horizontal road

    a) i) Car has negligible(?) resistance to its forward motion and is accelerating at 0.5 m/s2. Calculate the driving force
    Using F=ma, F=1500x0.5 = 750N

    ii) The car has a driving force of 2500N and total resistance to its forward motion of 250N. Calculate acceleration
    Again, F=ma, 2500-250=1500a, 1500a=2250, a=1.5m/s2

    b) Car of mass 1500kg is pulling 900kg trailer along straight horizontal road. The connection is light, rigid and horizontal. Driving force of 600N acting on car
    i) give the horizontal forces acting on the car and trailer seperately
    Car=> forward=600N, backwards=tension(T1) and friction (muR)=(mu1500g) would you include friction?
    Trailer=> forward=600N(?) and tension(T2), backwards=friction (muR)=(mu900g)

    ii) calculate the acceleration of the car and trailer. Also calculate the force in the coupling stating whether it is a tension of a thrust
    Im not too sure how to do this part, I know it would require F=ma but when I do it, it leaves too many unknowns

    iii) In a new situation, the trailer has a resistance to motion of 300N and the car has a resistance RN. There is no driving force and there is zero tension in the coupling calculate R

    2 - A girl of mass 48kg stands in a lift that is going upwards. The lift initially accelerates at 2m/s2 and then travels at a constant speed of 1.5m/s. Finally, the lift decelerates at 3m/s2. The normal reaction of the floor of the lift on the girl is RN

    a) draw a diagram of the weight of the girl and the normal reaction of the floor on her. Write down value of R when the lift is travelling at constant speed
    b) Calculate value of R when lift is accelerating at 2m/s2
    c) calculate value of R when lift is decelerating at 3m/s2

    The girl travels up in the lift on another occasion when she is holding a parcel of mass 5kg by means of a light inextendable string. The lift moves as before.

    d) the string does not break during the upward motion. What force must the string be able to sustain? calculate value of the normal reaction of the floor of the lift on the girl (holding the parcel) while the lift is accelerating

    I have no idea on these parts :/

    thanks if you made it through this far and any input would be much welcomed
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  2. #2
    Forum Admin topsquark's Avatar
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    Re: Mechanics - Dynamics

    Quote Originally Posted by adamcobz View Post
    Is this the right way to work out these answers?

    1 - Car has weight of 1500kg and is travelling along a straight, horizontal road

    a) i) Car has negligible(?) resistance to its forward motion and is accelerating at 0.5 m/s2. Calculate the driving force
    Using F=ma, F=1500x0.5 = 750N

    ii) The car has a driving force of 2500N and total resistance to its forward motion of 250N. Calculate acceleration
    Again, F=ma, 2500-250=1500a, 1500a=2250, a=1.5m/s2

    b) Car of mass 1500kg is pulling 900kg trailer along straight horizontal road. The connection is light, rigid and horizontal. Driving force of 600N acting on car
    i) give the horizontal forces acting on the car and trailer seperately
    Car=> forward=600N, backwards=tension(T1) and friction (muR)=(mu1500g) would you include friction?
    Trailer=> forward=600N(?) and tension(T2), backwards=friction (muR)=(mu900g)

    ii) calculate the acceleration of the car and trailer. Also calculate the force in the coupling stating whether it is a tension of a thrust
    Im not too sure how to do this part, I know it would require F=ma but when I do it, it leaves too many unknowns

    iii) In a new situation, the trailer has a resistance to motion of 300N and the car has a resistance RN. There is no driving force and there is zero tension in the coupling calculate R
    These are a bit long to put into one post. Next time please only put one of these per post.

    Best to do the first first and get to the second after this one is done.

    a)
    i) Good.

    ii) Good.

    b)
    i) Don't borrow trouble. If friction is not mentioned you can usually assume that it isn't there. Note also that the "line" between the car and trailer is ideal, that is to say that the tension on the car is the same as the tension on the trailer.

    ii) The tension in the connector cancels out when considering the car and trailer as the same system. So do your Newton's 2nd by considering the system as a single mass.

    iii) For there to be no tension in the connector then both car and trailer must have the same speed, non-zero in general. So if the trailer has a net force on it, what must the net force on the car be?

    Can you finish it?

    -Dan
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  3. #3
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    Re: Mechanics - Dynamics

    Okay thanks, I will do that next time

    for ii) would it simply leave one force of 600N pushing the system of 2400kg forward - making acceleration 1/4 m/s2

    as for iii) would the 'net force' (ie R) for the car simply be 300N, the same as the trailer?
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  4. #4
    Forum Admin topsquark's Avatar
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    Re: Mechanics - Dynamics

    Quote Originally Posted by adamcobz View Post
    Okay thanks, I will do that next time

    for ii) would it simply leave one force of 600N pushing the system of 2400kg forward - making acceleration 1/4 m/s2

    as for iii) would the 'net force' (ie R) for the car simply be 300N, the same as the trailer?
    Right on both counts.

    -Dan
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