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16 September, 11:44

A 1130-kg car is held in place by a light cable on a smooth (frictionless) ramp. The cable makes an angle of 31.0° above the surface of the ramp, and the ramp itself rises at 25.0° above the horizontal. Determine the tension in the cable?

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  1. 16 September, 12:04
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    T = 5163.89 N

    Explanation:

    Newton's first law:

    ∑F = 0 Formula (1)

    ∑F : algebraic sum of the forces in Newton (N)

    We define the x-axis in the direction parallel to the movement of the car on the ramp and the y-axis in the direction perpendicular to it.

    Forces acting on the car

    W: Weight of the car : In vertical direction

    FN : Normal force : perpendicular to the ramp

    T : Tension force: at angle of 31.0° above the surface of the ramp

    Calculated of the Weight of the car (W)

    W = m*g m: mass g:acceleration due to gravity

    W = 1130-kg * 9.8 m/s² = 11074 N

    x-y weight components

    Wx = 11074 N*sin 25.0° = 4680.07 N

    Wy = 11074 N*cos 25.0° = 10036.45 N

    x-y Tension components

    Tx = T*cos 25.0°

    Ty = T*sin 25.0°

    Newton's first law:

    ∑Fx = 0 Formula (1)

    Tx-Wx = 0

    T*cos 25.0° - 4680.07 = 0

    T*cos 25.0° = 4680.07

    T = 4680.07 / cos 25.0°

    T = 5163.89 N
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