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    Just a question for anyone who can answer it regarding M1:The in-extensible rod connecting the two bodies (Truck and car where the truck is the heavier body) is inclined at 30 degrees to the horizontal. If the system is decelerating at a speed of 1.02 ms^-2 , find the constant braking force on the heavier particle and the force in the rigid rod connecting both particles.The link to the diagram is below: http://www.thestudentroom.co.uk/atta...6&d=1368668728
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    (Original post by examkid)
    Just a question for anyone who can answer it regarding M1:The in-extensible rod connecting the two bodies (Truck and car where the truck is the heavier body) is inclined at 30 degrees to the horizontal. If the system is decelerating at a speed of 1.02 ms^-2 , find the constant braking force on the heavier particle and the force in the rigid rod connecting both particles.The link to the diagram is below: http://www.thestudentroom.co.uk/atta...6&d=1368668728
    What are the forces 300N and 400N shown on the diagram?
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    they are constant resistances to motion.
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    (Original post by examkid)
    Just a question for anyone who can answer it regarding M1:The in-extensible rod connecting the two bodies (Truck and car where the truck is the heavier body) is inclined at 30 degrees to the horizontal. If the system is decelerating at a speed of 1.02 ms^-2 , find the constant braking force on the heavier particle and the force in the rigid rod connecting both particles.The link to the diagram is below: http://www.thestudentroom.co.uk/atta...6&d=1368668728
    Did you draw the diagram yourself? It doesn't make much sense to me because the braking force and the resistance to motion can't be in the direction of motion.

    Constant braking force=
    B+400+300-(400+1600)x9.8xsin30=(400+1600)x 1.02
    =11140N






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    Guys.. it's decelerating. It's original motion was in the opposite direction to the constant braking force.
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    (Original post by examkid)
    Guys.. it's decelerating. It's original motion was in the opposite direction to the constant braking force.
    Have you got the answer to the question?
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    (Original post by freakynerdlol)
    ...
    take a pic of the question and upload....


    ryan
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    (Original post by brianeverit)
    What are the forces 300N and 400N shown on the diagram?
    In that case we will simply have
    for the truck  F+400+T\cos 30=1600\times 1.02 and for the car
    300-T\cos 30=400\times 1.02 so adding we have
     F+700=2000\times 1.06 \rightarrow F=1340 N
    or simply
    total external force on system is -(F+700)N so -(F+700)=-2000x1.02 giving same answer
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    (Original post by brianeverit)
    In that case we will simply have
    for the truck  F+400+T\cos 30=1600\times 1.02 and for the car
    300-T\cos 30=400\times 1.02 so adding we have
     F+700=2000\times 1.06 \rightarrow F=1340 N
    or simply
    total external force on system is -(F+700)N so -(F+700)=-2000x1.02 giving same answer
    Thank you. That was what I was looking for, I could not take a picture of the question as this question (more or less - I couldn't remember the values so I made them up) was in the M1 may 2013 international paper.
 
 
 
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