Fireman\'s acceleration down a fire-pole is -4.2m/s 2 . He weighs 97kg (mg) What

Question

Fireman's acceleration down a fire-pole is -4.2m/s2. He weighs 97kg (mg)

What is the upward force (F) exerted upward by the pole on the fireman

F= 97 (9.8 - 4.2) = 540N

My question is why is his acceleration down the pole subtracted from the acceleration of gravity? I get why 4.2 is negative. I just would think that since he is accelerating down at -4.2m/s2 and gravity is -9.8m/s2 that it would be more like 97 (-9.8 - 4.2). Looking at the equation it makes sense why gravity is not negative as a varible in this equation. It does not make sense to me in real life why the acceleration of gravity and someone accelerating downward dont get added together. I can do the problem and get it right because I follow the formula. When I was plugging the numbers in though it made me wonder.

Fireman's acceleration down a fire-pole is -4.2m/s2. He weighs 97kg (mg) What is the upward force (F) exerted upward by the pole on the fireman Using Newton's third law sigma F=ma Sigma Fy= F-mg right arrow F-mg=ma right arrow F=mg+ ma right arrow F=m(g+a) F= 97 (9.8 - 4.2) = 540N My question is why is his acceleration down the pole subtracted from the acceleration of gravity? I get why 4.2 is negative. I just would think that since he is accelerating down at -4.2m/s2 and gravity is -9.8m/s2 that it would be more like 97 (-9.8 - 4.2). Looking at the equation it makes sense why gravity is not negative as a varible in this equation. It does not make sense to me in real life why the acceleration of gravity and someone accelerating downward dont get added together. I can do the problem and get it right because I follow the formula. When I was plugging the numbers in though it made me wonder.

Explanation / Answer

Good Question.

It may be easy to think that if it was a free-fall, his acc would be = g in downward direction. But here it is lesser (4.2 m/s2 in downward direction). Hence pole must apply some force on him upwards so as to prevent him falling at g.

So, by Newton's Law,

Fgravity - Fby pole = ma

-mg + Fby pole = m(-4.2)

=> mg - Fby pole = m(4.2)

=> Fby pole = m(9.8-4.2) = 540N

Note: geff = g - a (where geff is in downward direction as you can see from sign of g... g & a are vectors)

You are wrong in saying that we are not taking g as negative. Why they are not added together? Bcz they are on other sides (mg is in force side and a in acc side (F=ma)).

Advise - Don't Just follow formulae... Go by the concept, else you will always face problem when you'll face any new problem.

Cheers

PS - Ask if you have any doubt.

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