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11 November, 06:25

A firefighter with a weight of 707 N slides down a vertical pole with an acceleration of 3.04 m/s2, directed downward. (a) What are the magnitude and direction of the vertical force (use up as the positive direction) exerted by the pole on the firefighter?

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  1. 11 November, 09:20
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    Fup=492.072N

    Explanation:

    Given required solution.

    W=707N Fup=? from newton's 2nd law we know that

    let g=10N/kg Fnet=ma ... eq1

    a=3.04m/s2 so the net force is the total of all

    forces. In other words

    Fnet=W-Fup ... eq2

    substitute eq1 from eq2 so it becomes

    ma=w-Fup ... substitute the given numbers

    before that we should have to find the mass of the fighter to compute Fup by using the following formula.

    w=mg m=w/g⇒707/10=70.7kg

    then continue with the above equation.

    ma=w-Fup use the negative sign in front of Fup because it acts opposite

    ma-w=-Fup take w to the left to get Fup

    w-ma=Fup divide both sides by negative one

    707N-70.7kg*3.04m/s2=Fup

    Fup = 707N-214.928N

    Fup=492.072N

    Note that the following;

    W is weight of fighter

    Fup is the upward reaction force.

    m is the mass of the fighter

    a is the acceleration of the fighter.
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