Incline plane, force problem

[dorkymichelle]

Homework Statement

A motorcycle and 60.6 kg rider accelerate at 3.5 m/s2 up a ramp inclined 7.3° above the horizontal. What are the magnitude of (a) the net force on the rider and (b) the force on the rider from the motorcycle?

Homework Equations

F=ma
Fg= mg

3. The Attempt at a Solution


I'm going to use F1 = force that is making the rider go up, which is exerted from the motorcycle ?
F2 is component of gravity that opposes the motorcycle from going up.

so netforce = F1-F2
F= ma
F1-F2=ma
F2 = sin 7.3degrees = y/mg
F2 = sin 7.3degrees = y/60.6*9.8
F2=75.46
using F1-F2 = ma
I got
F1-75.46 = 60.6*3.5
F1-75.46 = 212.1
F1 = 287.56
So force that motorcycle exerts on rider is 287.56 N, since that is what's making the rider and motorcycle go up.
I also think that the net force is 287.56 because you have gravity and normal force that cancels out, then you have F1 and F2 and F1-F2 = 287.56 N
I think where I went wrong is not putting in the mass of the motorcycle, but I'm not sure how to find that?

 

[tiny-tim]

welcome to pf!

hi dorkymichelle! welcome to pf!

(have a degree: ° and try using the X₂ icon just above the Reply box )

A motorcycle and 60.6 kg rider accelerate at 3.5 m/s2 up a ramp inclined 7.3° above the horizontal. What are the magnitude of (a) the net force on the rider and (b) the force on the rider from the motorcycle?

i think you're misunderstanding the question …

there are two forces on the rider, the weight mg, and the reaction force R from the motorcycle (which is the answer to (b)) …

the net force is the sum of those two forces (as vectors, of course)

 

[dorkymichelle]

Hmm.. what do you mean weight mg, the weight of the rider or the weight of the motorcycle?

 

[tiny-tim]

hi dorkymichelle!

(just got up :zzz: …)

there are two forces on the rider, the weight mg, and the reaction force R from the motorcycle (which is the answer to (b)) …

Hmm.. what do you mean weight mg, the weight of the rider or the weight of the motorcycle?

we're only considering the forces on the rider,

so it has to be the weight of the rider

(for F = ma, or for a vector triangle, or for a free body diagram, we always use all the forces on one body only)

 

[rwisz]

Your approach is on the right track, but there are a few errors in your calculations. Let's break it down step by step:

1. The first thing we need to do is identify all of the forces acting on the rider and motorcycle. These include:

- The force of gravity (Fg), which is pulling the rider and motorcycle down towards the ground.
- The normal force (Fn), which is the force exerted by the ramp on the motorcycle and rider in the perpendicular direction.
- The force of friction (Ff), which is the force that opposes motion between the motorcycle's tires and the ramp.
- The force of the motorcycle (Fm), which is the force that the motorcycle exerts on the rider in the direction of motion.
- The force of the rider (Fr), which is the force that the rider exerts on the motorcycle in the opposite direction.

2. Next, we need to draw a free body diagram to visualize all of these forces. This will help us see which forces cancel out and which ones we need to consider in our calculations.

3. Now, let's break down the forces into their components along the ramp. We can do this by using trigonometry and the angle of inclination (7.3°). The components are:

- The force of gravity (Fg) has a component along the ramp (Fgx) and a component perpendicular to the ramp (Fgy).
- The normal force (Fn) has a component along the ramp (Fnx) and a component perpendicular to the ramp (Fny).
- The force of friction (Ff) also has a component along the ramp (Ffx) and a component perpendicular to the ramp (Ffy).
- The force of the motorcycle (Fm) has a component along the ramp (Fmx) and a component perpendicular to the ramp (Fmy).
- The force of the rider (Fr) has a component along the ramp (Frx) and a component perpendicular to the ramp (Fry).

4. Now, let's write out the equations for each of these components using Newton's second law (F=ma). Remember that the net force on an object is equal to the sum of all the forces acting on it.

- Along the ramp (x-direction): Fnet = Fgx + Fnx + Ffx + Fmx + Frx = max
- Perpendicular to the ramp (y-direction): Fnet = Fgy + F