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2 February, 14:58

A 6.00-kg box sits on a ramp that is inclined at 37.0° above the horizontal. the coefficient of kinetic friction between the box and the ramp is mk = 0.30. what horizontal force is required to move the box up the incline with a constant acceleration of 3.60 m>s 2?

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  1. 2 February, 17:43
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    We need to see what forces act on the box:

    In the x direction:

    Fh-Ff-Gsinα=ma, where Fh is the horizontal force that is pulling the box up the incline, Ff is the force of friction, Gsinα is the horizontal component of the gravitational force, m is mass of the box and a is the acceleration of the box.

    In the y direction:

    N-Gcosα = 0, where N is the force of the ramp and Gcosα is the vertical component of the gravitational force.

    From N-Gcosα=0 we get:

    N=Gcosα, we will need this for the force of friction.

    Now to solve for Fh:

    Fh=ma + Ff + Gsinα,

    Ff=μN=μGcosα, this is the friction force where μ is the coefficient of friction. We put that into the equation for Fh.

    G=mg, where m is the mass of the box and g=9.81 m/s²

    Fh=ma + μmgcosα+mgsinα

    Now we plug in the numbers and get:

    Fh=6*3.6 + 0.3*6*9.81*0.8 + 6*9.81*0.6 = 21.6 + 14.1 + 35.3 = 71 N

    The horizontal force for pulling the body up the ramp needs to be Fh=71 N.
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