Dynamics and inclined planes
In a nutshell
To solve problems where there is motion on an inclined plane, you can use the same principles already seen in static equilibrium problems. However any motion in a particular plane can be considered by using F=ma in a particular direction. You can still use ∑F=0 for a plane where there is no motion as well as F=μR.
Equations
DESCRIPTION | EQUATION |
---|
When an object accelerates in a particular plane, use Newton's second law | F=m×a |
When an object does not accelerate in a particular plane, the resultant force is zero. | |
Friction | |
Variable definitions
QUANTITY NAME | SYMBOL | UNIT NAME | UNIT |
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Force of friction | | | |
Reaction force | | | |
| | | |
Acceleration | | Meters per second squared | |
Coefficient of friction | | no units | |
Dynamics and inclined planes
Procedure
1. | Draw a force diagram. |
2. | Resolve the forces into different components; refer to the inclined plane. |
3. | For the plane where there is no acceleration, use ∑F=0 to form equations. |
4. | For the plane where there is acceleration, use F=ma to form equations. |
5. | Solve the equations. |
Example 1
A fox of mass m slides from rest on a slide inclined at θ=52∘ to the horizontal. The fox accelerates down the slide at a=2.4 m.s−2. Calculate the coefficient of friction μ.
First draw a force diagram:
Resolve the forces into components, parallel and perpendicular to the plane. You should add labels to your diagram as follows:
Resolving perpendicular to the plane gives:
R=Wcos(52)R=mgcos(52)
Resolving parallel to the plane gives:
FP=Wsin(52)−Fmgsin(52)−μ×mgcos(52)=ma=ma=m×2.4
Solve for μ to give:
μμ=mgcos(52)mgsin(52)−2.4m=0.882 (3 s.f.)