Forces on Inclined Planes

Up to now, every object that has been sliding/moving against a surface has just been moving along a horizontal surface. However, not all surfaces are horizontal. The majority of surfaces in “real life” happen to have some incline (slant) to them. When we analyze forces on an incline, things are pretty different than analyzing forces on a flat, horizontal surface. The diagram below represents all the forces that might pop up when doing incline plane problems.

Motion of

The object Ff

F11

N

“the ground”

mg F

F11

Notice in the above drawing, the angle of the incline is the same angle the weight (mg) makes with a force called F(the perpendicular force---described below). This will always be true and is very important to solve any incline plane problems.

When an object is on an incline, the following forces always act on it:

mg= This is still the object’s weight and it still points down (directly down, toward the center of the earth).

F= “F-perpendicular”. This is the amount of the object’s weight that is perpendicular to the surface of the incline. To calculate it, use a trig function with the angle between F and the mg, the weight being the “hypotenuse” (see diagram above). The F┴ is usually the same as the N (provided there are no other forces in the same plane/direction that are pressing on/lifting the object).

F11= “F-parallel”. This is the amount of the object’s weight that is parallel to the surface of the incline; it is a force that is always directed down the incline. In order to calculate it, use a trig function and the angle between the mg and F. Notice that the F11 and the F are perpendicular with each other (together, the mg, the F, and the F11 make a right triangle).

N = This is still the normal force; it never equals the object’s total weight in incline plane problems. It still represents the amount of force that the surface of the incline is pushing back on the object with. In order to calculate it, you would need to find the total force that the object presses into the surface of the incline with (which will typically be equal the F┴).

Ff= This still represents the friction force that exists between the two surfaces moving against each other. It always opposes the motion of the object. If the object is sliding down the plane, the friction force will oppose it and point up the plane. If something is pushing/pulling the object up the plane, friction will oppose it and be directed down the plane.

Whenever you’re doing “forces on incline plane” problems, you need to remember that all of your vectors in the problem must be in the same “plane” when doing calculations. For example, if you’re trying to find the acceleration of the object down the plane, you would need to know only the forces that are acting on the object in that plane (i.e., the friction force, the parallel force, and any additional (external) applied force acting on the box in that plane that would act parallel to the acceleration of the object). The hardest part (which isn’t hard at all, trust me) is doing the initial calculations to find the parallel and perpendicular forces.

Here’s a problem: (Get ready for some fun….)

Two masses are connected by a rope. Mass one (20Kg) is on an incline plane (with a coefficient of friction of 0.15) that is at a 30-degree angle to the horizontal, and mass two (5Kg) is connected to mass 1 with a pulley and dangles in the air (until it is released, at which point it will move). When it’s released, mass 1 will move down the incline and mass 2 will move up in the air (I’m telling you this, you’re not required to be able to figure out which way it will move). What will their acceleration be? What is the tension in the rope?

20Kg

5Kg

300

Draw ALL the forces acting on BOTH masses:

T T

Ff

20Kg 5Kg

F11

N mg2

300 F┴

mg1

The above diagram is of the two masses, one is on an inclined plane and the other is suspended from a pulley. The masses are connected by the same cord, and the cord has the same amount of tension in it all the way through, so both masses will have the same acceleration.

Here’s what you know:

m1 = 20Kg

m2 = 5Kg

μ= 0.15

A problem like this will typically ask you two things: (1) What is the acceleration (and then you’ll have to do something with the acceleration)? (2) What is the tension in the rope? Since each mass is connected by the same rope, each acceleration will be the same (but opposite direction or signs) and the tension in the rope will be the same throughout. For this problem, I am going to say that the5Kg mass will risevertically upas the 20 kg mass slides down the incline.(The direction of object motion will usually be given to you…it’s not really an “eyeball and guess thing” that you’ll have to determine on your own). Since the 20 kg mass is sliding down the incline, friction will point up the incline.

In order to solve it, you’ll need to write the force equation for the entire mass of the system(Fnet = ma), just like in the “pulley” problems. The tension is pulling up on both masses, so it will cancel out and not affect the entire system acceleration. Since, again, the acceleration is directed “down” the incline, we need all the forces acting on the 20Kg mass that are also along the same direction/plane as the incline (parallel to the plane). Therefore, you will also need to solve for the amount of the mass’s weight that is pulling it down the incline (F11).

You will need to solve for the F11, the Ff, and the mg2 before you eventually get your final answer. Since the Ff = μ N, you will first need to find N. The normal force in incline problems is usually equal to the amount of the object’s weight that is directed perpendicular to the object’s motion (F┴)--look at the original diagram.

m1g = 196N

F11 = 98N

F┴ = 169.7N

N = 169.7N

Ff = 25.46N (found by Ff = μN)

m2g = 49N

a = ?

T = ?

Fnet = m2g + Ff - F11 = (m1 + m2)a

m2g + Ff - F11 = a

(m1 + m2)

a = -0.94 m/s/s

This negative acceleration means that the vector points down the incline so it will make mass 1 speed up as it moves down the incline and mass 2 speed up as it moves up in the air. Since the 20Kg mass is accelerating “down” the incline, its acceleration will be negative and the acceleration of the 5Kg mass will be positive since it’s moving up in the air.

To find the tension, plug the “a” in to the individual force equation for either Mass 2 or Mass 1 and you get:

Example:

For Mass 1T + Ff - F11 = m1(-a)

*Remember to use the negative acceleration as Mass 1 is accelerating down the incline

 T = m1(-a) + F11 - Ff

 T = 53.74 N

For Mass 2T - m2g = m2(+a)

 T = m2(+a) + m2g

T = 53.7 N

*This value for T makes sense as the tension needs to be greater than the weight for Mass 2 to get it to accelerate up.

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Unit 3, Part 3 Notes---Forces on Inclined Planes