How Much Force Can Be Applied Before a Penguin Slides Off a Sled?

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Homework Help Overview

The problem involves a sled being pulled horizontally with a penguin on it, focusing on the forces at play due to friction. The subject area includes concepts of kinetic and static friction, as well as forces acting on a system in motion.

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  • Mixed

Approaches and Questions Raised

  • Participants discuss the calculations of friction forces acting on both the sled and the penguin, exploring the relationship between the applied force and the frictional forces. Questions arise regarding the acceleration of the system and how it relates to the forces acting on the penguin.

Discussion Status

Some participants have provided calculations and attempted to derive expressions for acceleration in terms of the applied force. There is ongoing exploration of the conditions under which the penguin begins to slide, with no clear consensus on the correct approach yet.

Contextual Notes

Participants are working under the constraints of static and kinetic friction coefficients, and there is a noted urgency to resolve the problem within a limited time frame.

ace214
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A sled weighing 54.0 N is pulled horizontally across snow so that the coefficient of kinetic friction between sled and snow is 0.100. A penguin weighing 71.0 N rides on the sled, as in Fig. P4.76. If the coefficient of static friction between penguin and sled is 0.700, find the maximum horizontal force that can be exerted on the sled before the penguin begins to slide off.
p4-76.gif


I figured the friction on the sled with Fk = .1(71 N + 54 N) = 12.5 N
Static friction on the penguin is Fs <= .7(71 N) <= 49.7

So I figured that that the point where the penguin is just about to slip off is where the pulling force minus the kinetic friction equals the static friction of the penguin. So I did

F - fk = fs
F = fs + fk = 62.2

This isn't right however... Any help would be appreciated..
 

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What is the acceleration of the system in terms of F?

Examining the freebody diagram of the penguin... what is the force that keeps the penguin moving forward? Solve for this force in terms of F...

slipping occurs when the friction on the penguin reaches 49.7N.
 
learningphysics said:
What is the acceleration of the system in terms of F?

Examining the freebody diagram of the penguin... what is the force that keeps the penguin moving forward? Solve for this force in terms of F...

slipping occurs when the friction on the penguin reaches 49.7N.

The force moving the penguin forward is static friction...

Ok here's what I tried.
F - fk = ma
F = ma + fk where m = 125/9.8 = 12.8 and fk = 12.5
so aSystem = (F - 12.5)/12.8
and then aPenguin = (-F + 49.7)/7.24

I don't really know what to do with these because the a's aren't the same... Are they equal in magnitude but opposite in direction?
If you take them as the same and solve and then plug them back to find F you get the wrong answer.
I'm still stuck.
 
Bump... Need this solved tonight... :-(
 
You got the acceleration of the system is:

(F - 12.5)/12.8

This is also the acceleration of the penguin.

so using the freebody diagram of the penguin.

friction = ma

friction = (71/9.8)(F-12.5)/12.8

we know that friction < = 0.700(71) (this is just uk*mg)

hence (71/9.8)(F-12.5)/12.8 <= 0.700(71)

So what is the maximum value for F?
 
ace214 said:
and then aPenguin = (-F + 49.7)/7.24

this is where you're making the mistake. F is not acting on the penguin.
 

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