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30 March, 06:28

A sprinter is just coming out of the starting block, and only one foot is touching the block. The sprinter pushes back horizontally against the block with a force of 800 N and pushes down vertically against the ground with a force of 1000 N. How large is the resultant force and in what direction and angle is the resultant force located

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  1. 30 March, 06:51
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    Answer: 1280.63 N, 231.35°

    Explanation: Both the vertical and horizontal force are perpendicular to each other.

    The horizontal force is 800 N to the back (which corresponds to a force of 800 N acting in the negative x direction).

    The vertical force is 1000 N acting down vertically against the ground (which corresponds to a force of 1000 N acting in the negative y direction)

    Hence the resultant force is placed in the third quadrant (between the negative x and negative y axis).

    The magnitude of resultant force is gotten below as

    R = √ (Fx) ² + (Fy) 2

    Fx = 800 and Fy = 1000 N

    R = √800² + 1000²

    R = √640000 + 1000000

    R = √1640000 = 1280.63 N

    direction = tan^-1 (1000/800) = tan^-1 (1.25) = 51. 35°

    But the force is on the third quadrant, hence angle is 180 + 51.35° = 231.35°
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