Energy/Work Problem: Friction of sliding down pole

In summary, the problem involves a fireman of mass m sliding down a pole with distance d. He starts from rest and reaches a constant speed at the bottom, as if he had stepped off a platform a distance h≤d above the ground. The task is to find the average friction force exerted on the pole in terms of the given variables m, d, h, and appropriate constants. By setting up a conservation equation at the bottom and at height h with no friction, the average friction force can be calculated as: mg(d-h)/(d-h) = F, where F is the average friction force.
  • #1
BBA Biochemistry

Homework Statement


A fireman of mass m slides a distance d down a pole. He starts from rest. He moves as fast at the bottom as if he had stepped off a platform a distance h≤d above the ground and descended with negligible air resistance.What average friction force did the fireman exert on the pole?
Express your answer in terms of the variables m, d, h, and appropriate constants.

Homework Equations


U=mgh
K= 0.5mv^2
K1+U1+W(other)=K2+U2
W=Fs

The Attempt at a Solution


Since friction is a nonconservative force, I figure I'm supposed to use K1+U1+W(other)=K2+U2 and find the W(other), and then use the equation W=Fs to find the actual force of the friction.

When the fireman is at rest at height, d, he has no KE, just PE, which is:

U1= mgd

As he reaches the ground, some of the potential energy was converted to kinetic energy, but the rest of it was lost through friction:

mgd+W(other)=K2=0.5mv^2

To find W(other) I need to find the velocity of the fireman. This is where I start to get unsure. The problem states that his velocity is constant at all heights (h) after he starts sliding down from his initial position, d. I decided to set up another conservation equation, in which he starts at some height, h, and then reaches the ground:

mgh=0.5mv^2 - W(other2)

Since W=Fs, and s=h, I substituted:

mgh=0.5mv^2 - Fh

Where F is the friction force I want to find out. I solved for v^2, getting:

v^2= (2/m) * (mgh + Fh)

I plugged this back into the conservation equation for the entire distance, getting:

W(other)=mgd-0.5m[(2/m) * (mgh + Fh)]=mgd-mgh+Fh

W(other) is also equal to a force over a distance, so:

mgd-mgh+Fh=W=Fd

Isolating F:

mg(d-h)=Fd-Fh=F(d-h)

Then the (d-h) terms cancel, getting mg=F. However, this is incorrect, and the problem states that the answer has the term d in it.

Can anyone point out where my logic and/or math was wrong?
 
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  • #2
BBA Biochemistry said:
As he reaches the ground, some of the potential energy was converted to kinetic energy, but the rest of it was lost through friction:

mgd+W(other)=K2=0.5mv^2
You went off the rails (a bit) when you introduced the speed "v". Instead, just express U2+ K2 in terms of the given variables.

Also, the work done by friction is negative, so be careful with signs.
 
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  • #3
I'm not really sure how to express U2+K2 in terms of the given variables d, h, and m.

I'm at mgd = U1 and K1=0.

So mgd + 0 + W(other) = U2 + K2.

Can I equate W(other)=F*s, making the equation:

mgd + F*d = U2 + K2 ?

If U2 equals zero (since the fireman is on the ground), that leaves:

mgd + F*d = K2.

But I still don't know K2 or F. Any hints on where I'm supposed to get my second equation for a system of equations? I haven't used the "He moves as fast at the bottom as if he had stepped off a platform a distance h≤d above the ground and descended with negligible air resistance." part of the problem yet.
 
  • #4
BBA Biochemistry said:
mgd + F*d = K2.
Good. Now find the corresponding equation for the case where there is no friction and the starting height is h.
 
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  • #5
haruspex said:
Now find the corresponding equation for the case where there is no friction and the starting height is h.

Would that be as simple as:

mgh = K2

Since there is no initial kinetic energy nor final potential energy?

Since speed is the same at the bottom regardless of air resistance (according to the problem), would I just set these two equations equal to each other and solve for Fk?
 
  • #6
I just submitted the answer and this was correct. Thanks for the help everyone!
 

Related to Energy/Work Problem: Friction of sliding down pole

1. What is friction?

Friction is a force that resists the motion between two surfaces in contact. It is caused by irregularities on the surfaces that come into contact with each other.

2. How does friction affect the sliding down of a pole?

Friction acts in the opposite direction of the motion, so it will slow down the sliding down of a pole. The amount of friction depends on the materials of the pole and the surface it is sliding on, as well as the weight and speed of the person sliding down.

3. How can we calculate the friction force when sliding down a pole?

To calculate the friction force, we need to know the coefficient of friction, which is a measure of how easily the two surfaces slide against each other. We also need to know the normal force, which is the force exerted by the surface on the object due to its weight. The friction force can be calculated by multiplying the coefficient of friction by the normal force.

4. How can we reduce the friction when sliding down a pole?

To reduce friction when sliding down a pole, we can use a lubricant such as oil or wax on the surface of the pole. This will create a smoother surface and decrease the coefficient of friction. Additionally, reducing the weight or speed of the person sliding down can also decrease the friction force.

5. What other factors can affect the friction when sliding down a pole?

The roughness of the surface and the temperature can also affect the friction when sliding down a pole. Rough surfaces will have a higher coefficient of friction, while smoother surfaces will have a lower coefficient of friction. Additionally, an increase in temperature can decrease the coefficient of friction, making it easier to slide down the pole.

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