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?

Fup=492.072N

Explanation:

Given                         required                   solution.

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

let g=10N/kg                                  Fnet=maeq1      

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

                                             forces.In other words  

                                      Fnet=W-Fupeq2

substitute eq1 from eq2 so it becomes

ma=w-Fupsubstitute 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.