Place a book on a tilted surface and it either slides down or stays put depending on the angle and how rough the surface is. What determines this? This resource helps you to understand the components of gravitational forces that affect this and use them to analyse forces on inclined surfaces. Also learn about how an understanding of forces on slopes can be applied, like in the Budj Bim aquaculture system.
When an object sits on a slope, gravity doesn't act along or against the surface directly; it acts straight down. To analyse the motion, we break that gravitational force into two components:
the component perpendicular to the slope, which pushes the object into the surface and produces the normal force and friction
the component parallel to the slope, which drives the object down the incline.
Breaking a force into components like this is called resolving the force.
Just as we can analyse horizontal and vertical motion separately on a flat surface, here we analyse forces perpendicular and parallel to the sloping surface. The same principle applies: the normal force always acts perpendicular to the surface, while friction and the driving force act parallel to it.
Resolving the components of gravitational force
Consider this object on a plane, where weight (\(W=mg\)) acts through the centre of mass.
Forces parallel to the slope
A frictional force \(F_{F}\) acts on the object, stopping it from sliding down the slope. The force driving the object down the slope is given by \(mg\sin\theta\). The difference between these two forces is the resultant force down the slope:
\[\sum F = mg\sin\theta -F_{F}\]
Friction acts to oppose sliding motion, preventing the object from sliding down the plane. For example, if the mass were being dragged uphill, friction would act downhill.
Forces perpendicular to the slope
The normal force \(F_{N}\) is at right angles to the surface. The force pushing the object into the surface is given by \(mg\cos\theta\). These forces are equal and opposite one another, otherwise the object would fall through the plane. The resultant force perpendicular to the slope is zero because it sits on the slope, hence:
\[mg\cos\theta=F_{N}\]
To work out the angles, remember that sum of angles in any triangle is \(180^{\circ}\) and a right angle is \(90^{\circ}\).
Indigenous knowledges in physics
The Budj Bim aquaculture complex
The steeper the slope, the faster something moves down it. The Gunditjmara people of South-Western Victoria understood and applied the principle across more than 6,600 years of aquaculture at Budj Bim, a UNESCO World Heritage Site on Gunditjmara Country.
Using volcanic rock from the ancient Budj Bim lava flows, the Gunditjmara constructed an intricate network of channels, weirs, dams and holding ponds to manage the seasonal flow of water from nearby Lake Condah. The gradient of each channel determined the speed of water flow through it: too steep and kooyang (short-finned eels) would be swept through rather than trapped; too shallow and water would not flow at all.
Budj Bim, image by Mertie via Flickr, licensed under CC BY 2.0
This knowledge has never been static. Through colonisation, the Gunditjmara were dispossessed of their Country, forced onto missions and actively discouraged from their traditional practices. Following severe flooding in 1946, the government constructed a large drain through Lake Condah—without consulting the Gunditjmara—that left the lake dry by the late 1950s, silencing more than 6,600 years of active aquaculture.
In 2008, Lake Condah was reclaimed by the Gunditjmara people and in 2010, a cultural weir was constructed, reinstating traditional water flows across the aquaculture system. The restoration was led by Gunditjmara Traditional Owners, with Indigenous workers involved in every aspect of the construction. In 2019, the Budj Bim Cultural Landscape was inscribed on the UNESCO World Heritage List – the first site in Australia to receive this recognition purely for its Aboriginal cultural importance.
Daniell KA and Moggridge B (2024) 'Indigenous Water Engineering and Aquaculture Systems in Australia: The Budj Bim Cultural Landscape and Baiame's Ngunnhu (the Brewarrina Aboriginal Fish Traps)', Blue Papers, 3(1):18–29, doi:10.58981/bluepapers.2024.1.01
A toy car of mass \(\mathbf{50}\,\textbf{g}\) travels down a smooth incline at \(\mathbf{30^{\circ}}\) to the horizontal. Calculate:
the net force acting on the car as it rolls down the slope
Always start with a force diagram. Friction may be ignored as the incline is described as 'smooth'. Gravity, \(g=9.8\,\text{m s}^{-2}\). As the slope is \(30^{\circ}\), then \(90-30=60^{\circ}\) in top corner and \(90-60=30^{\circ}\) from the normal to vertical force. This allows us to use \(mg\sin30^{\circ}\) as the component parallel to the slope.
As \(mg\sin30^{\circ}\) is the component of the force parallel to the slope:
\[\begin{align*} \sum F & = ma \\[6pt]
& = mg\sin30^{\circ}-F_{f}\\[6pt]
& = mg\sin30^{\circ}-0\\[6pt]
& = mg\sin30^{\circ} \end{align*}\]
The surface is frictionless (\(F_{F}=0\)). Therefore, the only force allowing the car to roll down the incline is the component of the gravitational force \(mg\sin\theta\). Remember to convert mass in \(g\) to \(kg\).
The net force acting on the car as it rolls down the slope is \(0.25\,\text{N}\).
the force of the incline on the car as it travels down the slope.
Again, as the slope is \(30^{\circ}\), the angle in the top corner is \(90-30=60^{\circ}\) and the angle from the normal to vertical force is \(90-60=30^{\circ}\). This allows us to use \(mg\cos30^{\circ}\) as the component perpendicular.
The force of the incline on the car \(mg\cos\theta\) is equal to the normal force \(F_{N}\).
The force of the incline on the car as it travels down the slope is \(0.43\,\text{N}\).
The steepest road in the world is Baldwin Street in Dunedin, New Zealand. It has an incline of \(\mathbf{52^{\circ}}\). Ignoring friction, how quickly would a car left with its handbrake off accelerate down this street? Baldwin Street in Dunedin, New Zealand, by Tristan Schmurr via Flickr, licensed under CC BY 2.0
Since the surface is frictionless, the only force acting on the slope is \(mg\sin\theta\). Applying Newton's second law:
A car left with its handbrake off would accelerate at \(7.7\,\text{m s}^{-2}\) down Baldwin Street if the road was frictionless.
Exercise – calculating forces on slopes
A skateboarder riding a skateboard of total mass \(60\,\text{kg}\) coasts down a frictionless ramp at an angle of \(30^{\circ}\) to the horizontal. Remember, \(g=9.8\,\text{m s}^{-2}\).
Calculate the normal force acting on the rider and skateboard.
Calculate the force acting on the rider and skateboard parallel to the ramp.
\(509\,\text{N}\)
\(294\,\text{N}\)
The skateboarder now coasts down another ramp with the same angle, but this time the ramp has a rough surface.
Calculate the normal force acting on the rider and skateboard.
Calculate the force acting on the rider and skateboard parallel to the ramp due to gravity.
If the ramp has a frictional force of \(54\,\text{N}\), what is the net force acting on the rider and the skateboard?
Calculate the acceleration of the skateboarder.
If the skateboarder started from rest and the ramp is \(4\,\text{m}\) long, what was the speed of the skateboarder at the bottom of the ramp?
