If a 68kg person climbs a rope at .61m/s in 10 sec, how is Force calculated?

Question

I know F = m*a but it also is equal to m*g. In this case, m is equal to 68kg and "a" is found by .61/10s. But on the other hand, would it be 68kg * 9.8m/s2?

Answer 1

This seems like the type of question where they would ask for work done. This is just a guess, but asking for the force is not logical given the known values. To calculate the work, you need:
Work = Force*distance*cos(theta)
First find the distance:
Distance = V*t
Distance = 6.1 meters
Force upward will be equal to the force of gravity downward due to Newton's third law. So in this case F = MG
F = 666.4N
The cosine of the angle will be zero because the person is moving in the same direction as the force. Cos(0) = 1, so we leave that out.
Work done = 666.4*6.1 = 4065.04 Newton meters or jewels.
Again, I am sorry if this is not what you are looking for, but it is the only logical thing.

Answer 2

With each pull, there is a spurt of acceleration which can be added to gravity, then multiplied by the person's mass. As the climber progresses, the force is not constant, but it is usually mass*gravity except for the spurts upward followed by a drop as he reaches up for the next pull.

So if you wanted to know the force, you'd have to look at each phase of what is happeneing. While there is not much going on in the force column, there is a lot or work going on (force*distance).

Answer 3

Looks to me like he is climbing at a constant speed of .61 m/s. Not accelerating, so there is no force other than 68*9.8 Newtons.

Answer 4

sigma F=ma and Fg=mg. He is accelerating upwards, to the sigma F is in that direction. Do a force diagram and a vector diagram and you will realize that Fg= Fguy + Sigma F. Therefore, Fg-Sigma F = Fguy. So mg-ma = Fguy. 68(10) - 68(.061) = 675.852 N. Remember: g is not equal to a, that is only in falling problems.

Answer 5

in my opinion, make it alot easier and use vectors