Inclined plane

An inclined plane, also known as a ramp, is a flat supporting surface tilted at an angle from the

Moving an object up an inclined plane requires less

The

Two other simple machines are often considered to be derived from the inclined plane.

The term may also refer to a specific implementation; a straight ramp cut into a steep hillside for transporting goods up and down the hill. This may include cars on rails or pulled up by a cable system; a funicular or cable railway, such as the Johnstown Inclined Plane.

Uses

Inclined planes are widely used in the form of loading ramps to load and unload goods on trucks, ships and planes.

.

Other inclined planes are built into permanent structures. Roads for vehicles and railroads have inclined planes in the form of gradual slopes, ramps, and

.

History

Stevin's proof

In 1586, Flemish engineer Simon Stevin (Stevinus) derived the mechanical advantage of the inclined plane by an argument that used a string of beads.[10] He imagined two inclined planes of equal height but different slopes, placed back-to-back as in a prism (A, B, C above). A loop of string with beads at equal intervals is draped over the inclined planes, with part of the string hanging down below. The beads resting on the planes act as loads on the planes, held up by the tension force in the string at point T. Stevin's argument goes like this:[10][11][12] The string must be stationary, in static equilibrium. If the string was heavier on one side than the other, and began to slide right or left under its own weight, when each bead had moved to the position of the previous bead the string would be indistinguishable from its initial position and therefore would continue to be unbalanced and slide. This argument could be repeated indefinitely, resulting in a circular perpetual motion , which is absurd. Therefore, it is stationary, with the forces on the two sides at point T (above) equal. The portion of the chain hanging below the inclined planes is symmetrical, with an equal number of beads on each side. It exerts an equal force on each side of the string. Therefore, this portion of the string can be cut off at the edges of the planes (points S and V), leaving only the beads resting on the inclined planes, and this remaining portion will still be in static equilibrium. Since the beads are at equal intervals on the string, the total number of beads supported by each plane, the total load, is proportional to the length of the plane. Since the input supporting force, the tension in the string, is the same for both, the mechanical advantage of each plane is proportional to its slant length As pointed out by Dijksterhuis,[13] Stevin's argument is not completely tight. The forces exerted by the hanging part of the chain need not be symmetrical because the hanging part need not retain its shape when let go. Even if the chain is released with a zero angular momentum, motion including oscillations is possible unless the chain is initially in its equilibrium configuration, a supposition which would make the argument circular.

Inclined planes have been used by people since prehistoric times to move heavy objects.^[14][15] The sloping roads and causeways built by ancient civilizations such as the Romans are examples of early inclined planes that have survived, and show that they understood the value of this device for moving things uphill. The heavy stones used in ancient stone structures such as Stonehenge^[16] are believed to have been moved and set in place using inclined planes made of earth,^[17] although it is hard to find evidence of such temporary building ramps. The Egyptian pyramids were constructed using inclined planes,^[18][19][20] Siege ramps enabled ancient armies to surmount fortress walls. The ancient Greeks constructed a paved ramp 6 km (3.7 miles) long, the Diolkos, to drag ships overland across the Isthmus of Corinth.^[4]

However the inclined plane was the last of the six classic

It was not until the

The first elementary rules of sliding

Terminology

Slope

The mechanical advantage of an inclined plane depends on its slope, meaning its gradient or steepness. The smaller the slope, the larger the mechanical advantage, and the smaller the force needed to raise a given weight. A plane's slope s is equal to the difference in height between its two ends, or "rise", divided by its horizontal length, or "run".^[31] It can also be expressed by the angle the plane makes with the horizontal, ${\displaystyle \theta }$.

${\displaystyle \theta =\tan ^{-1}{\bigg (}{\frac {\text{Rise}}{\text{Run}}}{\bigg )}\,}$

Mechanical advantage

The mechanical advantage ${\displaystyle \mathrm {MA} }$ of a simple machine is defined as the ratio of the output force exerted on the load to the input force applied. For the inclined plane the output load force is just the gravitational force of the load object on the plane, its weight ${\displaystyle F_{\text{w}}}$. The input force is the force ${\displaystyle F_{\text{i}}}$ exerted on the object, parallel to the plane, to move it up the plane. The mechanical advantage is

${\displaystyle \mathrm {MA} ={\frac {F_{\text{w}}}{F_{\text{i}}}}\,}$

The ${\displaystyle \mathrm {MA} }$ of an ideal inclined plane without friction is sometimes called ideal mechanical advantage ${\displaystyle \mathrm {IMA} }$ while the MA when friction is included is called the actual mechanical advantage ${\displaystyle \mathrm {AMA} }$.^[32]

Frictionless inclined plane

If there is no friction between the object being moved and the plane, the device is called an ideal inclined plane. This condition might be approached if the object is rolling like a barrel, or supported on wheels or casters. Due to conservation of energy, for a frictionless inclined plane the work done on the load lifting it, ${\displaystyle W_{\text{out}}}$, is equal to the work done by the input force, ${\displaystyle W_{\text{in}}}$^[33][34][35]

${\displaystyle W_{\rm {out}}=W_{\rm {in}}\,}$

Work is defined as the force multiplied by the displacement an object moves. The work done on the load is equal to its weight multiplied by the vertical displacement it rises, which is the "rise" of the inclined plane

${\displaystyle W_{\rm {out}}=F_{\rm {w}}\cdot {\text{Rise}}\,}$

The input work is equal to the force ${\displaystyle F_{\text{i}}}$ on the object times the diagonal length of the inclined plane.

${\displaystyle W_{\rm {in}}=F_{\rm {i}}\cdot {\text{Length}}\,}$

Substituting these values into the conservation of energy equation above and rearranging

${\displaystyle {\text{MA}}={\frac {F_{\rm {w}}}{F_{\rm {i}}}}={\frac {\text{Length}}{\text{Rise}}}\,}$

To express the mechanical advantage by the angle ${\displaystyle \theta }$ of the plane,^[34] it can be seen from the diagram (above) that

${\displaystyle \sin \theta ={\frac {\text{Rise}}{\text{Length}}}\,}$

So

${\displaystyle {\text{MA}}={\frac {F_{\rm {w}}}{F_{\rm {i}}}}={\frac {1}{\sin \theta }}\,}$

So the mechanical advantage of a frictionless inclined plane is equal to the reciprocal of the sine of the slope angle. The input force ${\displaystyle F_{\rm {i}}}$ from this equation is the force needed to hold the load motionless on the inclined plane, or push it up at a constant velocity. If the input force is greater than this, the load will accelerate up the plane. If the force is less, it will accelerate down the plane.

Inclined plane with friction

Where there is friction between the plane and the load, as for example with a heavy box being slid up a ramp, some of the work applied by the input force is dissipated as heat by friction, ${\displaystyle W_{\text{fric}}}$, so less work is done on the load. Due to conservation of energy, the sum of the output work and the frictional energy losses is equal to the input work

${\displaystyle W_{\text{in}}=W_{\text{fric}}+W_{\text{out}}\,}$

Therefore, more input force is required, and the mechanical advantage is lower, than if friction were not present. With friction, the load will only move if the net force parallel to the surface is greater than the frictional force ${\displaystyle F_{\text{f}}}$ opposing it.^[8][36][37] The maximum friction force is given by

${\displaystyle F_{f}=\mu F_{n}\,}$

where ${\displaystyle F_{\text{n}}}$ is the normal force between the load and the plane, directed normal to the surface, and ${\displaystyle \mu }$ is the

between the two surfaces, which varies with the material. When no input force is applied, if the inclination angle ${\displaystyle \theta }$ of the plane is less than some maximum value ${\displaystyle \phi }$ the component of gravitational force parallel to the plane will be too small to overcome friction, and the load will remain motionless. This angle is called the of the angle of repose ${\displaystyle \phi }$ is equal to ${\displaystyle \mu }$

${\displaystyle \phi =\tan ^{-1}\mu \,}$

With friction, there is always some range of input force ${\displaystyle F_{\text{i}}}$ for which the load is stationary, neither sliding up or down the plane, whereas with a frictionless inclined plane there is only one particular value of input force for which the load is stationary.

Analysis

A load resting on an inclined plane, when considered as a free body has three forces acting on it:^[8][36][37]

• The applied force, ${\displaystyle F_{\text{i}}}$ exerted on the load to move it, which acts parallel to the inclined plane.
• The weight of the load, ${\displaystyle F_{\text{w}}}$, which acts vertically downwards
• The force of the plane on the load. This can be resolved into two components:
  • The normal force ${\displaystyle F_{\text{n}}}$ of the inclined plane on the load, supporting it. This is directed perpendicular (
    normal
    ) to the surface.
  • The frictional force, ${\displaystyle F_{\text{f}}}$ of the plane on the load acts parallel to the surface, and is always in a direction opposite to the motion of the object. It is equal to the normal force multiplied by the
    coefficient of static friction
    μ between the two surfaces.

Using

the load will be stationary or in steady motion if the sum of the forces on it is zero. Since the direction of the frictional force is opposite for the case of uphill and downhill motion, these two cases must be considered separately:

• Uphill motion: The total force on the load is toward the uphill side, so the frictional force is directed down the plane, opposing the input force.

Derivation of mechanical advantage for uphill motion The equilibrium equations for forces parallel and perpendicular to the plane are ${\displaystyle \sum F_{\|}=F_{i}-F_{f}-F_{w}\sin \theta =0\,}$ ${\displaystyle \sum F_{\perp }=F_{n}-F_{w}\cos \theta =0\,}$ Substituting ${\displaystyle F_{f}=\mu F_{n}\,}$ into first equation ${\displaystyle F_{i}-\mu F_{n}-F_{w}\sin \theta =0\,}$ Solving second equation to get ${\displaystyle F_{n}=F_{w}\cos \theta \,}$ and substituting into the above equation ${\displaystyle F_{i}-\mu F_{w}\cos \theta -F_{w}\sin \theta =0\,}$ ${\displaystyle {\frac {F_{w}}{F_{i}}}={\frac {1}{\sin \theta +\mu \cos \theta }}\,}$ Defining ${\displaystyle \mu =\tan \phi \,}$ ${\displaystyle {\frac {F_{w}}{F_{i}}}={\frac {1}{\sin \theta +\tan \phi \cos \theta }}={\dfrac {1}{\sin \theta +{\dfrac {\sin \phi }{\cos \phi }}\cos \theta }}={\frac {\cos \phi }{\sin \theta \cos \phi +\cos \theta \sin \phi }}\,}$ Using a sum-of-angles trigonometric identity on the denominator,

The mechanical advantage is

${\displaystyle \mathrm {MA} ={\frac {F_{w}}{F_{i}}}={\frac {\cos \phi }{\sin(\theta +\phi )}}\,}$

where ${\displaystyle \phi =\tan ^{-1}\mu \,}$. This is the condition for impending motion up the inclined plane. If the applied force F_i is greater than given by this equation, the load will move up the plane.

• Downhill motion: The total force on the load is toward the downhill side, so the frictional force is directed up the plane.

Derivation of mechanical advantage for downhill motion The equilibrium equations are ${\displaystyle \sum F_{\|}=F_{i}+F_{f}-F_{w}\sin \theta =0\,}$ ${\displaystyle \sum F_{\perp }=F_{n}-F_{w}\cos \theta =0\,}$ Substituting ${\displaystyle F_{f}=\mu F_{n}\,}$ into first equation ${\displaystyle F_{i}+\mu F_{n}-F_{w}\sin \theta =0\,}$ Solving second equation to get ${\displaystyle F_{n}=F_{w}\cos \theta \,}$ and substituting into the above equation ${\displaystyle F_{i}+\mu F_{w}\cos \theta -F_{w}\sin \theta =0\,}$ ${\displaystyle {\frac {F_{w}}{F_{i}}}={\frac {1}{\sin \theta -\mu \cos \theta }}\,}$ Substituting in ${\displaystyle \mu =\tan \phi \,}$ and simplifying as above ${\displaystyle {\frac {F_{w}}{F_{i}}}={\frac {\cos \phi }{\sin \theta \cos \phi -\cos \theta \sin \phi }}\,}$ Using another trigonometric identity on the denominator,

The mechanical advantage is

${\displaystyle \mathrm {MA} ={\frac {F_{w}}{F_{i}}}={\frac {\cos \phi }{\sin(\theta -\phi )}}\,}$

This is the condition for impending motion down the plane; if the applied force F_i is less than given in this equation, the load will slide down the plane. There are three cases:

1. ${\displaystyle \theta <\phi \,}$: The mechanical advantage is negative. In the absence of applied force the load will remain motionless, and requires some negative (downhill) applied force to slide down.
2. ${\displaystyle \theta =\phi \,}$: The 'angle of repose'. The mechanical advantage is infinite. With no applied force, load will not slide, but the slightest negative (downhill) force will cause it to slide.
3. ${\displaystyle \theta >\phi \,}$: The mechanical advantage is positive. In the absence of applied force the load will slide down the plane, and requires some positive (uphill) force to hold it motionless

Mechanical advantage using power

The mechanical advantage of an inclined plane is the ratio of the weight of the load on the ramp to the force required to pull it up the ramp. If energy is not dissipated or stored in the movement of the load, then this mechanical advantage can be computed from the dimensions of the ramp.

In order to show this, let the position r of a rail car on along the ramp with an angle, θ, above the horizontal be given by

${\displaystyle \mathbf {r} =R(\cos \theta ,\sin \theta ),}$

where R is the distance along the ramp. The velocity of the car up the ramp is now

${\displaystyle \mathbf {v} =V(\cos \theta ,\sin \theta ).}$

Because there are no losses, the power used by force F to move the load up the ramp equals the power out, which is the vertical lift of the weight W of the load.

The input power pulling the car up the ramp is given by

${\displaystyle P_{\mathrm {in} }=FV,\!}$

and the power out is

${\displaystyle P_{\mathrm {out} }=\mathbf {W} \cdot \mathbf {v} =(0,W)\cdot V(\cos \theta ,\sin \theta )=WV\sin \theta .}$

Equate the power in to the power out to obtain the mechanical advantage as

${\displaystyle \mathrm {MA} ={\frac {W}{F}}={\frac {1}{\sin \theta }}.}$

The mechanical advantage of an inclined plane can also be calculated from the ratio of length of the ramp L to its height H, because the sine of the angle of the ramp is given by

${\displaystyle \sin \theta ={\frac {H}{L}},}$

therefore,

${\displaystyle \mathrm {MA} ={\frac {W}{F}}={\frac {L}{H}}.}$

Example: If the height of a ramp is H = 1 meter and its length is L = 5 meters, then the mechanical advantage is

${\displaystyle \mathrm {MA} ={\frac {W}{F}}=5,}$

which means that a 20 lb force will lift a 100 lb load.

The Liverpool Minard inclined plane has the dimensions 1804 meters by 37.50 meters, which provides a mechanical advantage of

${\displaystyle \mathrm {MA} ={\frac {W}{F}}=1804/37.50=48.1,}$

so a 100 lb tension force on the cable will lift a 4810 lb load. The grade of this incline is 2%, which means the angle θ is small enough that sin θ≈tan θ.

See also

References

External links

Wikimedia Commons has media related to Inclined planes.

• An interactive simulation of Physics inclined plane