Inclined Planes
A complete guide for the IMAT Exam
1. Resolving Gravity
The most important step for any inclined plane problem is to handle the force of gravity, \(F_g = mg\). Gravity always acts straight down. However, it's more useful to resolve this force into two components:
- Parallel Component (\(F_{g\parallel}\)): Acts parallel to the surface of the incline. This is the component that tries to pull the object down the slope.
- Perpendicular Component (\(F_{g\perp}\)): Acts perpendicular to the surface of the incline. This component pushes the object into the surface.
\(F_{g\parallel} = mg \sin(\theta)\)
\(F_{g\perp} = mg \cos(\theta)\)
Where \(\theta\) is the angle of the incline.
2. The Normal Force (\(F_N\))
The Normal Force is the support force exerted by the surface on the object. It is always perpendicular to the surface. On an inclined plane, the object is not accelerating in the perpendicular direction. Therefore, the Normal Force must balance the perpendicular component of gravity.
\(F_N = F_{g\perp} = mg \cos(\theta)\)
Common Mistake: The Normal Force is NOT equal to \(mg\) on an incline!
3. Friction (\(F_f\))
Friction is a force that opposes motion or attempted motion. Its magnitude depends on the Normal Force and the coefficient of friction (\(\mu\)).
- Static Friction (\(F_s\)): Acts on objects at rest. It has a maximum value it can reach before the object starts to slide. The object remains at rest if \(F_{g\parallel} \le F_{s,max}\).
- Kinetic Friction (\(F_k\)): Acts on objects that are already in motion. It's usually less than the maximum static friction.
Maximum Static Friction: \(F_{s,max} = \mu_s F_N = \mu_s mg \cos(\theta)\)
Kinetic Friction: \(F_k = \mu_k F_N = \mu_k mg \cos(\theta)\)
Example 1: Will it Slide?
A 10 kg box rests on a ramp inclined at 30°. The coefficient of static friction \(\mu_s\) is 0.6. Will the box slide down the ramp? (Use \(g = 9.8 \, m/s^2\))
Step 1: Calculate the force pulling the box down the ramp (Parallel Force).
\(F_{g\parallel} = mg \sin(\theta) = (10 \, \text{kg})(9.8 \, m/s^2) \sin(30^\circ) = (98)(0.5) = 49 \, \text{N}\)
Step 2: Calculate the maximum available static friction force.
First, find the Normal Force: \(F_N = mg \cos(\theta) = (10)(9.8) \cos(30^\circ) \approx (98)(0.866) = 84.87 \, \text{N}\)
Now, find max static friction: \(F_{s,max} = \mu_s F_N = (0.6)(84.87 \, \text{N}) = 50.92 \, \text{N}\)
Step 3: Compare the forces.
The pulling force (\(49 \, \text{N}\)) is less than the maximum friction force (\(50.92 \, \text{N}\)).
Conclusion: The box will not slide. The force of static friction will match the parallel force and be exactly 49 N.
Example 2: Calculating Acceleration
A 5 kg crate slides down a 37° incline. The coefficient of kinetic friction \(\mu_k\) is 0.3. What is the acceleration of the crate? (Use \(g = 10 \, m/s^2\), \(\sin(37^\circ) \approx 0.6\), \(\cos(37^\circ) \approx 0.8\))
Step 1: Identify all forces acting parallel to the incline.
The net force (\(F_{net}\)) is the difference between the parallel component of gravity and the kinetic friction force. \(F_{net} = F_{g\parallel} - F_k\)
Step 2: Calculate these forces.
\(F_{g\parallel} = mg \sin(\theta) = (5)(10)(0.6) = 30 \, \text{N}\)
\(F_N = mg \cos(\theta) = (5)(10)(0.8) = 40 \, \text{N}\)
\(F_k = \mu_k F_N = (0.3)(40 \, \text{N}) = 12 \, \text{N}\)
Step 3: Calculate the net force.
\(F_{net} = 30 \, \text{N} - 12 \, \text{N} = 18 \, \text{N}\)
Step 4: Use Newton's Second Law (\(F_{net} = ma\)) to find acceleration.
\(18 \, \text{N} = (5 \, \text{kg}) a\)
\(a = \frac{18}{5} = 3.6 \, m/s^2\)
Conclusion: The crate accelerates down the incline at \(3.6 \, m/s^2\).