Chimney climber, Newton's lawquestion

  • Thread starter hemetite
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  • #1
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Hi, i just need to see whether my answer is logically correct?

A 70kg climber is supported in the chimney by the friction forces exerted on his shoes and back. The static coefficients of frictions between his shoes and the wall, and between his back and the wall, are 0.80 and 0.60 respectively. What is the minimum force he must exert? Assume the walls are vertical and that friction forces are both at maximum...

Here is my answer.

For the climber to remain in contact with the wall.

Both Frictional Force(shoe) and Frictinal Force(back) = mg = 70*9.81

else he will fall...

Fn1= the force he must exert by his shoe
Fn2= the force he must exert by his back

therefore

Frictional Force(shoe)= 0.80 * Fn1

mg=0.80*Fn1

Fn1(shoe)=853.4N

Frictional Force(back) = 0.60* Fn2
mg=0.6*Fn2
Fn2(back)=1144.5N.

Therefore minimum normal force he must exert to satisfy both...= 1144.5N

Logical and correct?
 

Answers and Replies

  • #2
HallsofIvy
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Why must he support his total weight with his feet and support his total weight with his back? Do you see where Newton's Law comes in? The force he exerts on the walls of the chimney with his feet and back are equal.
 
  • #3
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I knew both force need to be equal

you are saying that i should dived the mg into 2...

50% load handle by the shoe

50% load handle by his back..?
 
  • #4
Doc Al
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For the climber to remain in contact with the wall.

Both Frictional Force(shoe) and Frictinal Force(back) = mg = 70*9.81
No. As Halls' already explained, don't set each individual friction force equal to the weight. Instead, the total upward force must equal his weight.
else he will fall...

Fn1= the force he must exert by his shoe
Fn2= the force he must exert by his back
How are these forces related? Hint: Analyze the horizontal forces acting on the climber.
I knew both force need to be equal

you are saying that i should dived the mg into 2...

50% load handle by the shoe

50% load handle by his back..?
No. Don't make any assumptions about the friction forces being equal.
 
  • #5
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Okay..then it should be like this

Fn=exertion force

0.80*Fn + 0.6*Fn = mg

Fn = (70*9.81) / (0.80+0.60) = 686.7 / 1.4 = 490.5N

correct?
 
  • #6
Doc Al
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Looks good.
 
  • #7
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thanks.. :)
 

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