# The Normal Force and Inclined Planes

1. Jan 30, 2008

### faunmia

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
a= gsin(angle)
F=ma

3. 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

2. Jan 30, 2008

### andrevdh

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. Jan 30, 2008

### 07:45am

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: Jan 30, 2008
4. Jan 31, 2008

### andrevdh

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)

$$\Sigma F_x = 0$$
therefore
$$0 = 450 - Ff - mg\sin(5^o)$$
or
$$Ff = 450 - mg\sin(5^o)$$

Last edited: Jan 31, 2008
5. Jan 31, 2008

### 07:45am

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: Jan 31, 2008
6. Jan 31, 2008

### andrevdh

Do you understand the solution? It seems you got it from somewhere else?

7. Jan 31, 2008

### 07:45am

Oh yes, I understand it... I study Engineering Technology and last year we studied all these types of questions.