When dealing with resolving forces on inclined planes, follow these steps:

1. Draw a free body diagram: Show all forces acting on the object, including weight, normal force, friction (if present), and any applied force.
2. Resolve forces into components: Break the forces into two perpendicular directions:

• Parallel to the plane: The component of weight is $mg \sin(\theta)$, where $\theta$ is the angle of inclination.
• Perpendicular to the plane: The component of weight is $mg \cos(\theta)$. The normal force acts in this direction.

3. Apply $F = ma$: For equilibrium or motion along the plane, apply Newton's second law separately to the parallel and perpendicular directions. Use $\sum F = ma$ to find acceleration, tension, or friction. By resolving forces carefully and applying Newton's laws, you can solve for the forces acting on an object on an inclined plane.

Example Problem:

Problem: Find the resultant force and its direction for the following situation:

• Forces acting:
• 5 N upward
• 12 N downward
• 6 N to the right

Solution:

Step 1: Consider the forces in the horizontal and vertical directions.

• Horizontal Force (Rightward):

• Vertical Force (Net Force Downward):

Step 2: Calculate the resultant force F_R.

• Use the Pythagorean theorem:

Step 3: Find the direction of the resultant force.

• The angle $\alpha$ with the horizontal (rightward) is given by:

• Since $F_y$ is downward, the direction of the resultant force $F_R$ relative to the positive x-axis is:

Example 1: Finding the Values of F and R for a Particle at Rest on a Slope

Problem Statement:

• A particle of weight $5$g is held at rest on a slope inclined at $30^\circ$.
• Find the values of the force $F$ and the normal reaction $R$.

Solution:

1. Forces acting on the particle:

• The gravitational force acts downward: $5$g.
• $F$ is the force parallel to the slope.
• $R$ is the normal reaction force perpendicular to the slope.

2. Resolve forces along the slope:

• The component of weight along the slope: $5g \sin 30^\circ$.
• Since the particle is at rest, $F = 5g \sin 30^\circ$.

4. Resolve forces perpendicular to the slope:

• The component of weight perpendicular to the slope: $5g \cos 30^\circ$.
• Since the particle is at rest, $R = 5g \cos 30^\circ$.

Answer:

• $F = \frac{5g}{2} \, \text{N}$
• $R = \frac{5\sqrt{3}}{2} g \, \text{N}$

Example 2: Finding the Magnitudes of P and Q in Equilibrium

Problem Statement:

• A system is in equilibrium on a slope with forces $P$ and $Q$ acting on a particle.
• Find the magnitudes of $P$ and $Q$.

Solution:

5. Forces acting:

• The particle is subject to a normal force $2\cos 60^\circ$, a frictional force $6\sin 60^\circ$, and additional forces $P$ and $Q$.

6. Resolve forces along the slope:

• Horizontal components:

• Solving for $Q$:

7. Resolve forces perpendicular to the slope:

• Vertical components:

• Solving for $P$:

Answer:

• $P = 3 + \sqrt{3} \, \text{N}$
• $Q =-1 + 3\sqrt{3}N$

Problem Statement:

• A force $F$ is applied to a particle of mass $5 \, \text{kg}$, which holds it at rest on a smooth slope inclined at $30^\circ$ above the horizontal.
• Calculate the force $F$.

Problem Statement:

• A mass of $3$ kg rests on a smooth plane inclined at $45^\circ$ to the horizontal. The mass is attached to a cable that passes up the plane along the line of greatest slope and then passes over a smooth pulley at the top of the plane.
• The other end of the cable carries a mass of $1$ kg freely suspended.
• The masses are modelled as particles, and the cable as a light inextensible string
• There is a force $P N$ acting horizontally on the $3$ kg mass, and the system is in equilibrium. Calculate:

1. The magnitude of $P$.
2. The normal reaction between the mass and the plane.
3. State how the assumption that the pulley is smooth was used in the calculations.

8. Align axes with the plane: Make the plane one axis (usually parallel and perpendicular to the slope). This simplifies resolving forces into components like $mg \sin(\theta)$ and $mg \cos(\theta)$
9. Use clear trigonometry: Always resolve the weight into parallel ($mg \sin(\theta)$) and perpendicular ($mg \cos(\theta)$) components relative to the inclined surface.
10. Check for friction: If friction is present, remember it opposes motion and acts parallel to the plane. Use $F_{\text{friction}} = \mu F_{\text{normal}}$, where $\mu$ is the coefficient of friction and $F_{\text{normal}} = mg \cos(\theta).$