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The Normal Force and Inclined Planes

  • Thread starter faunmia
  • Start date
  • #1
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Homework Statement


A woman pushes her 150kg motorcycle up a slope of 5 degrees with a constant speed of 2 m/s. She achieves this by exerting a force on the bike of 450N parallel to the slope. What is the magnitude of the frictional force acting on the bike?


Homework Equations


a= gsin(angle)
F=ma

The Attempt at a Solution



a= (150 x 9.8) x sin5
F= 150 x 128
F= 19216.84

but the answer is ment to be 322 N, im not realy sure where I am going wrong, any help would be greatly appriciated
 

Answers and Replies

  • #2
andrevdh
Homework Helper
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The motorcycle is pushed at a constant speed up the slope. This means that the forces (components) acting along the direction of the slope cancel each other out.
 
  • #3
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ΣFx = 450N – Ff – 150kgx9.8ms-1xSin5
Ff = 321.8810582 = 322N


Or you could use the formula:
Ff = μFn

but you arent given μ...
 
Last edited:
  • #4
andrevdh
Homework Helper
2,128
116
Spot on with your first way of solving for Ff. It seems that you might be a bit unsure about getting to the solution due to the -1 in front of the sin?

We know that the sum of the forces along the slope need to be zero (no acceleration along the slope)

[tex]\Sigma F_x = 0[/tex]
therefore
[tex]0 = 450 - Ff - mg\sin(5^o)[/tex]
or
[tex]Ff = 450 - mg\sin(5^o)[/tex]
 
Last edited:
  • #5
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Ahh yes, sorry.. I made a mistake.

While calculating the Fn, I mistaken the g (in Fn=mg) to be a velocity... what i was supposed to write was:

150kg x 9.8ms^-2 x Sin5

Sorry for the confusion.
 
Last edited:
  • #6
andrevdh
Homework Helper
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116
Do you understand the solution? It seems you got it from somewhere else?
 
  • #7
3
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Oh yes, I understand it... I study Engineering Technology and last year we studied all these types of questions.
 

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