Inclined Plane Force Calculator
Calculate the force needed to push an object up an incline at a constant velocity.
Understanding the Forces on an Incline
The Downhill Forces
When an object is on a ramp, two forces work together to pull it downhill:
1. Parallel Gravity ($F_{g, \|}$): The component of gravity that acts parallel to the ramp's surface.
2. Friction ($F_f$): The resistive force between the object and the surface, which also points down the ramp when you are pushing up.
The Uphill Battle
To push the object up the ramp at a constant velocity, you must apply a force ($F_{app}$) that is exactly equal to the sum of the two downhill forces. Any less, and it won't move or will slide back. Any more, and it will accelerate up the ramp. This calculator solves for that perfect balancing force.
Interactive Diagram
Calculator
Derivation: How the Formula Works
The applied force is found by summing the two opposing forces: the parallel component of gravity and the kinetic friction force.
1 Sum the Opposing Forces
The total force you need to apply ($F_{app}$) is the sum of the parallel gravity component ($F_{g, \|}$) and the kinetic friction force ($F_{f,k}$).
2 Substitute Force Formulas
We substitute the formulas for each force: $F_{g, \|} = mg \sin(\theta)$ and $F_{f,k} = \mu_k F_N$. The Normal Force ($F_N$) on an incline is $mg \cos(\theta)$.
3 Final Equation
This gives us the final equation. Unlike the angle of repose, this calculation does depend on the mass of the object, as a heavier object requires more force to lift and creates more friction.
$$ F_{app} = mg(\sin(\theta) + \mu_k \cos(\theta)) $$
Common Questions & Examples
Example 1: Pushing a Box
A 20 kg box is on a ramp angled at 25°. The coefficient of kinetic friction is 0.4. What force is required to push it up the ramp at a constant velocity? (Assume g = 9.8 m/s²).
Solution:
- Parallel Gravity: $F_{g, \|} = 20 \times 9.8 \times \sin(25°) = 196 \times 0.423 = 82.8 N$
- Friction Force: $F_{f,k} = \mu_k \times (mg \cos(\theta)) = 0.4 \times (196 \times \cos(25°)) = 0.4 \times 177.6 = 71.1 N$
- Total Force: $F_{app} = 82.8 N + 71.1 N = 153.9 N$
- Result: $F_{app} \approx 153.9 N$
Example 2: Frictionless Plane
What force is needed to push a 5 kg block up a frictionless 45° incline?
Solution:
- Since the plane is frictionless, $\mu_k = 0$. The friction term becomes zero.
- The formula simplifies to: $F_{app} = mg \sin(\theta)$
- Substitute values: $F_{app} = 5 \times 9.8 \times \sin(45°) = 49 \times 0.707$
- Result: $F_{app} \approx 34.6 N$
What's the difference between static and kinetic friction?
Static Friction ($\mu_s$) is the force that prevents an object from starting to move. It's a variable force that matches the applied force up to a maximum value. It's generally stronger because it takes more force to overcome the initial "stickiness" between two surfaces at rest.
Kinetic Friction ($\mu_k$) is the force that opposes an object that is already in motion. It's typically a constant, lower value than the maximum static friction. This is why it's easier to keep a heavy box sliding than it is to get it moving in the first place.