Physics

A 6.0kg box is on a frictionless 45∘ slope and is connected via a massless string over a massless, frictionless pulley to a hanging 2.4kg weight.

What is the tension in the string if the 6.0kg box is held in place, so that it cannot move?

If the box is then released, which way will it move on the slope?

What is the tension in the string once the box begins to move?

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  1. 1.)

    If the box doesn't move, then the Tension in the rope is equal to Fg of the 2.4Kg mass.

    Fm1=m*a=Fg-T

    a=0m/s^2

    m*0m/s^2=(2.4kg*9.8m/s^2)-T

    T=Fg=(2.4kg*9.8m/s^2)=23.5N

    2.)

    Fm1=m1*a=T-Fg

    2.4kg*a=T-23.52N

    Fm2=m2*a=Fd-T

    Fd=mg*Sin45=(6.0kg*9.8m/s^2)*(0.70711)

    Fd=41.58N

    Fd>Fg for m1, so the box will move down the ramp, not up.

    3.)

    Fm2=m2*a=Fd-T

    Fd=41.58N

    Fm2=6.0kg*a=41.58N-T

    and

    Fm1=m1*a=T-Fg

    Fm1=2.4kg*a=T-23.52

    Solve for a:

    a=(T-23.52)/2.4kg

    Plug into equation for m2 and solve for T:

    6.0kg*[(T-23.52)/2.4kg]=41.58N-T

    (6.0T-141.12)/2.4kg=41.58N-T

    2.5T-58.8=41.58N

    2.5T=41.58N+58.8N

    2.5T=41.58N+58.8N

    2.5T=100.38

    T=40.2N

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  2. ERROR IS PREVIOUS ANSWER RESOLVED HERE
    In the third part of Devron's equations they dropped the minus T in going from

    (6.0T-141.12)/2.4kg=41.58N-T

    to

    2.5T-58.8=41.58N

    So it should be 3.5T-58.8=41.58N

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