Toggle Main Menu Toggle Search

Inclined planes(Mechanics)

A particle on an inclined plane

To solve problems about particles on inclined planes we need to resolve the forces parallel and perpendicular to the plane.

If a particle of mass $m$ is placed on a smooth inclined plane (i.e. the frictional force $F=0$) and released it will slide down the slope. To find the acceleration of the particle as it slides we resolve in the direction of motion. \begin{align} F & = ma, \\ mg \ \cos (90^{\circ} - \theta) & = ma, \\ g \ \cos (90^{\circ} - \theta) & = a, \\ g \ \sin (\theta) & = a. \end{align} Therefore, as the $m$'s cancel we can see that the particle's mass does not affect the acceleration - only the angle of the slope does.

If a particle of mass $m$ is placed on a rough inclined plane (i.e. the frictional force $F \neq 0$), the particle may be prevented from sliding if $F$ is large enough.

We resolve perpendicular to the plane (as the normal reaction $R$ acts at a right angle to the plane), where acceleration is zero. \begin{align} F & = ma, \\ R - mg \ \cos \theta & = m \times 0, \\ R & = mg \ \cos \theta. \end{align} Now we resolve in the direction of the slope, if the particle is at rest then \begin{align} F & = ma, \\ mg \ \cos (90^{\circ} - \theta) - F & = m \times 0, \\ mg \ \sin \theta & = F. \end{align} where $F$ is the force of friction as shown in the diagram. We know that the maximum frictional force is given by $F_{\text{MAX} } = \mu R$ therefore \begin{align} F & \leq \mu R, \\ mg \ \sin \theta & \leq \mu mg \ \cos \theta, \\ \frac{ \sin \theta}{ \cos \theta} & \leq \mu, \\ \tan \theta & \leq \mu. \end{align} Therefore the particle will remain at rest until $\tan \theta > \mu$, at this point it will accelerate down the slope.

Worked Example: Particle held at rest on an inclined plane

Finding the acceleration

A particle of mass $4 \mathrm{kg}$ is held at rest on a rough surface which is inclined to the horizontal at an angle $\theta$, where $\tan \theta = \frac{3}{4}$. The coefficient of friction between the particle and the surface is $\mu=0.3$. Find the acceleration of the particle when it is released.

Solution

You can draw a diagram similar to the one in the section above. As the particle is moving ($\tan \alpha \geq \mu$) the friction will be at it's maximum value $F_{\text{MAX } } = \mu R$ and acts in the opposite direction to acceleration (which acts down the slope as the particle slides).

We first resolve $F = ma$ in the direction perpendicular to acceleration to find the normal reaction, $R$. \begin{align} F & = ma, \\ R - mg \ \cos \theta &= m \times 0, \\ R & = mg \ \cos \theta, \\ & = 4 \times 9.8 \times 0.8, \\ & = 31.36 \mathrm{N}. \end{align} Note that because $\tan \theta = \frac{3}{4}$ we have that $\cos \theta = \frac{4}{5} = 0.8$ and $\sin \theta = \frac{3}{5} = 0.6$.

We can then resolve $F = ma$ in the direction of motion, which is down the slope. \begin{align} F & = ma, \\ mg \ \sin \theta - \mu R & = ma, \\ \left(4 \times 9.8 \times 0.6 \right) - \left( 0.3 \times 31.36 \right) & = 4a, \\ \frac{23.52 - 9.408}{4}& = a,\\ a& = 3.528\mathrm{ms^{-2} }. \end{align} The acceleration of the particle is $3.528 \mathrm{ms^{-2} }$.

Worked Example: Particle pushed up an inclined plane

Finding the acceleration

A particle of mass $3 \mathrm{kg}$ is pushed up a rough surface which is inclined to the horizontal at an angle $\theta = 25^{\circ}$. The coefficient of friction between the particle and the surface is $\mu=0.15$ and the magnitude of the horizontal force that pushes the particle is $30 \mathrm{N}$. Find the acceleration of the particle.

Solution

You can draw a diagram showing all the forces and the acceleration.

Note that $\tan 0.25^{\circ} > \mu$ so the particle is moving and therefore the friction is limiting. Acceleration is acting upwards as the particle is being pushed up the slope. We first resolve $F = ma$ in the direction perpendicular to acceleration to find the normal reaction, $R$. \begin{align} F & = ma, \\ R - mg \ \cos \theta - 30 \ \cos (90^{\circ} - \theta) &= m \times 0, \\ R & = mg \ \cos (25^{\circ}) + 30 \ \cos (90^{\circ} - 25), \\ & = (3 \times 9.8 \times 0.906) + (30 \times 0.423), \\ & = 39.324 \mathrm{N} \text{ (to 3d.p.).} \end{align}

We can then resolve $F = ma$ in the direction of motion, which is up the slope. \begin{align} F & = ma, \\ 30 \ \cos (25^{\circ}) - 3g \ \cos (65^{\circ}) - \mu R & = 3a, \\ 27.189 - 12.425 - (0.15 \times 39.324) & = 3 a, \\ a & = 2.955 \mathrm{ms^{-2} } \text{ (to 3d.p.).} \end{align} The acceleration of the particle is $2.955\mathrm{ms^{-2} }$ up the slope.

Test Yourself

Try our Numbas test on inclined planes: Inclined Plane

Whiteboard maths

More Support

You can get one-to-one support from Maths-Aid.