Learning Lab

Forces acting on a block on an inclined plane:

(At rest or slipping down the plane.)

Weight Force acting on a block resolved into its components:

Perpendicular and parallel to the incline. Remember that the normal force \(F_{N}\) is equal and opposite to the force exerted by the object on the plane \(mg\cos\theta\) (perpendicular to plane) otherwise it would fall through the plane, and \(mg\sin\theta\) (parallel to plane) forcing the object down the plane if no friction occurs.

To work out the angles remember sum of angles in any triangle is 180 degrees and remember a right angle is 90 degrees.

Note that with the above diagrams:

• Weight \(=mg\); acts through the centre of mass.

• Normal force \(F_{N}\) is always at right angles to the surface.

• Friction acts to oppose sliding motion (eg, if the mass were being dragged uphill, friction would act downhill)

• The weight force is resolved into 2 components:

• \(\ \ (1)\ \ \) perpendicular to plane, and

• \(\ \ (2)\ \ \) parallel to the plane.

• The resultant force \(\Sigma F\) down the slope is given by \(\Sigma F=mg\sin\theta-F_{f}\) where \(F_{f}\) is friction

• The resultant force \(\Sigma F\) perpendicular to the slope is zero (because it sits on the slope), hence: \(mg\cos\theta=F_{N}\)

Forces parallel to slope:

Note the angles: If slope is \(30\) degrees then \(90-30=60\) degrees in top corner then again \(90-60=30\) degrees from normal to vertical force so we would use \(mg\sin30\) down the slope as the sum of all forces.

(a)\(\ \ \ \ \ \) As \(mg\sin\theta\) is the component of the force parallel to the slope then “sum of all forces” = \(\Sigma F\):

\[\begin{align*} \Sigma F & =ma\\ & =mg\sin\theta-F_{f}\\ & =mg\sin\theta-0\\ & =mg\sin\theta \end{align*}\]

Where \(mg\sin\theta\) is the component of the force parallel to the slope.

Note: the surface is friction-less (smooth) ie. \(F_{f}=0\), therefore the only force allowing the car to roll down the incline is the component of the gravitational force ‘\(mg\sin\theta\)’ .

\[\begin{align*} \Sigma F & =mg\sin\theta\\ & =m\times g\times\sin\theta\\ & =50\times10^{-3}\times9.8\times\sin30\\ & =0.25N \end{align*}\]

Note: grams have been converted into kilograms

Forces perpendicular to slope:

Also note the angles: If slope is \(30\) degrees then \(90-30=60\) degrees in top corner then again \(90-60=30\) degrees from normal to vertical force so we use \(mg\cos30\) perpendicular to the slope as the normal force.

(b)\(\ \ \ \ \ \) The force of the incline on the car is a force that acts perpendicular to the slope, ie. the normal force \(F_{N}\) is equal to \(mg\cos\theta\)

\[\begin{align*} F_{N} & =mg\cos\theta\\ & =m\times g\times\cos\theta\\ & =50\times10^{-3}\times9.8\times\cos30\\ & =0.43N \end{align*}\]