Calculating Vertical Force on a Fireman Sliding Down a Pole

Click For Summary
To calculate the vertical force exerted by the pole on a fireman sliding down, it's essential to consider both the gravitational force and the net force due to acceleration. The fireman has a weight of 980N (100kg x 9.8m/s²) acting downward. While the net force from his acceleration is 300N upward, the total force exerted by the pole must counteract both the weight and provide the net force, resulting in a total of 680N upward. The correct approach involves using the equation ∑F = ma to account for all forces acting on the fireman. Understanding these forces clarifies the discrepancy in the calculations.
Amelina Yoo
Messages
14
Reaction score
0

Homework Statement


Fireman weighs 100kg. He slides down a vertical pole with acceleration 3.0ms^-2.

a) Magnitude and direction of vertical force exerted by the pole on fireman?

Homework Equations


f=ma

The Attempt at a Solution


[/B]
f=ma
f=100(3.0)
f=300N vertically up.

But the answer says that it is actually 680N vertically up, and I do understand how they came to this conclusion.
Can you please explain where to go beyond this point? Or point out a mistake I may have made? Thank you.

I also replaced with acceleration with gravity, which did not work.
 
Physics news on Phys.org
Your relevant equation is more correctly stated as:
∑F=ma

That is, the sum of all forces are equal to ma.
You found the net force from the firemans mass and acceleration, but you aren't asked for the net force, you are asked only for the force the pole exerts on the man.

What other force acts on the fireman?
What is it's magnitude? Can you use this information to solve for the poles force?
 
Amelina Yoo said:
f=100(3.0)
f=300N vertically up
And yet, that acceleration is vertically down. Strange, don't you think?
 
  • Like
Likes billy_joule
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 1 ·
Replies
1
Views
6K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 5 ·
Replies
5
Views
4K
Replies
4
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 8 ·
Replies
8
Views
11K
Replies
9
Views
2K