The Normal Force and Inclined Planes

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Homework Help Overview

The problem involves a woman pushing a motorcycle up an inclined plane with a specific angle and speed, seeking to determine the frictional force acting on the motorcycle. The subject area pertains to forces on inclined planes and friction in physics.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the balance of forces acting on the motorcycle, particularly the relationship between the applied force, frictional force, and gravitational component along the slope. Some express uncertainty regarding the calculations and assumptions made in determining the frictional force.

Discussion Status

Participants are actively engaging with the problem, offering different approaches to calculate the frictional force. Some have provided equations and reasoning, while others have questioned the assumptions made in the calculations. There is an acknowledgment of a mistake in one participant's approach, leading to further clarification and exploration of the concepts involved.

Contextual Notes

There is a mention of missing information regarding the coefficient of friction, which is not provided in the problem statement. Additionally, the discussion reflects on the implications of constant speed and the need for forces to balance out.

faunmia
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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, I am not really sure where I am going wrong, any help would be greatly appriciated
 
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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.
 
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:
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:
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:
Do you understand the solution? It seems you got it from somewhere else?
 
Oh yes, I understand it... I study Engineering Technology and last year we studied all these types of questions.
 

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