Imagine you had to load 50 barrels of root beer onto a truck, each weighing 250 pounds (117 kg). It would be an enormously difficult task to lift them up directly. But what if you could roll the barrels up a long ramp? If the slope were gentle enough, the job would be much easier.

## Redirection of Force

An inclined plane is a simple machine; a basic tool that helps us by redirecting or amplifying force. In this activity you will learn how an inclined plane redirects force.

1. On the SIMULATION pane of the Gizmo™, a brick lies halfway up an inclined plane. Check that the Coeff. of friction (μ) is set to 0.00, and click Play ().
1. How long did it take the brick to slide down the ramp? What was the Velocity of the brick at the bottom?
2. Click Reset (). Experiment with the Gizmo to find the relative effects of Weight and Angle on the motion of the brick. Which factor has no effect at all?
2. Select the FREE-BODY DIAGRAM tab, and check that the Magnitude mode is chosen. This diagram shows the forces on the brick. The Weight of the brick is a force that points straight downward, but the inclined plane breaks this force into two components: one parallel to the plane (W||) and one perpendicular (W).
1. Which force (W|| or W) will cause the brick to slide down the plane?
2. What is the ratio of W|| to the Weight of the brick? What is the ratio of W to the Weight?
3. Use a calculator to find the sine, cosine, and tangent of the Angle of the inclined plane. Do any of these resemble the ratios you just calculated?
4. Use this relationship to derive a formula for W|| and W in terms of weight (W) and Angle (θ). Test your formula with a variety of weights and angles.
3. Since the brick pushes downward on the plane, the plane must push upward on the brick. This is called the normal force and is represented by a red arrow on the diagram.
1. How does the magnitude of the normal force relate to W?
2. Given the directions and magnitudes of W, W||, and the normal force, what force represents the total net force on the brick?
3. Select the SIMULATION tab and note the Mass of the brick. Use the formula F = ma to predict what the acceleration (a) of the brick should be.
4. Click Play. Select the TABLE tab and look at the acceleration column. Was your predicted value correct? (Note: Because the brick is accelerating downwards, acceleration is negative.)
4. Click Reset and select the CONTROLS tab. Set the Coeff. of friction (μ) to 0.25. (To quickly set a slider to a certain value, type the value in the field to the right of the slider and press Enter.) Set the Angle to 30°, and select the FREE-BODY DIAGRAM tab.
1. The light blue arrow on the diagram represents the force of friction. How is the direction of this force related to the direction of movement?
2. What is the ratio between the force of friction and the value of W? How does this relate to the coefficient of friction, μ?
3. Write a formula for the force of friction based on μ and W. Use this formula to predict the force of friction when the Weight is 100 N, the Angle is 30°, and the Coeff. of friction (μ) = 0.50. Use the Gizmo to check your answer. (Note: The force of friction can never exceed W||. At most they will be equal.)
4. Based on the current force of friction, the mass of the brick, and W||, predict the acceleration of the brick. Use the Gizmo to check your answer.
5. Challenge: Write a formula for the acceleration of the brick based on its mass (m), the angle (θ), and the coefficient of friction (μ). (Recall that the weight of the brick is equal to the product of its mass and gravitational acceleration, 9.82 m/s2.)

In the previous activity, you found that ramps split the weight of an object into two forces, one of which is cancelled by the normal force. In this activity you will find how this property makes them useful for lifting objects.

1. Click Reset. Set the Angle to 48°, the Coeff. of friction (μ) to 0.00, and the Weight to 300 N. Change the External force to On, and a toy car appears to push the brick up the ramp. Check that the Applied force is 100 N and click Play.
1. Does the car apply enough force to push the brick uphill?
2. Click Reset. Increase the Applied force until you find a force sufficient to push the brick up the plane. What is this force?
3. Select the FREE-BODY DIAGRAM tab. How does this Applied force relate to the value of W||?
4. How much force would have been required if a forklift were used to lift the object directly, without the use of a ramp? Why does it take less force to lift the object using an inclined plane?
2. An inclined plane can make it easier to lift an object by changing the amount of force that must be overcome to move it. The mechanical advantage (MA) of an inclined plane is defined as

1. Use the results from the last exercise to create a formula for mechanical advantage in terms of weight (W) and W||.
2. In the first activity, you found the formula for W|| in terms of W and the angle, θ. Substitute this formula into your equation to create an equation for mechanical advantage in terms of just the angle θ.
3. Click Reset, set the Coeff. of friction (μ) to 0.20, and find the minimum Applied force required to push the brick up the ramp in this case. How does this force compare to the force of friction (displayed on the FREE-BODY DIAGRAM tab) plus W||? How does this affect the mechanical advantage of the inclined plane?
3. The car that pushes the brick up the ramp does work on the brick. If the car pushes the brick up the ramp by exerting a force F along a distance of d meters, the amount of work done is W = Fd. Set the Angle to 42°, the Coeff. of friction (μ) to 0.00, the Weight to 100 N, and the Applied force to 200 N. On the SIMULATION pane, drag the brick to a height of 0.6 m. Click Play.
1. Select the WORK tab. What was the distance the car pushed the brick?
2. Based on the Applied force and the distance, how much work was done on the brick?
4. As it was pushed up the ramp, the energy of the brick increased. This energy is expressed as potential energy and kinetic energy.
1. Subtract the initial height of the brick from the final height of the brick. How much height did the brick gain?
2. Potential energy in this case is equal to weight times height. How much potential energy did the brick gain as it was lifted?
3. What was the final velocity of the brick?
4. Kinetic energy is equal to 1/2 mv2. What was the final kinetic energy of the brick? (Remember to use the mass of the brick in this calculation.)
5. How much total energy did the brick gain? How does this compare to the amount of work done on the brick?
5. When there is no friction, an inclined plane is an ideal machine. When friction is present, some energy is wasted. The efficiency of a machine is the percentage of input work that "gets through" to the object. One way to measure this is to divide the amount of energy the object gained by the work done.
1. Based on the last simulation, what is the efficiency of an inclined plane when there is no friction? (Note: Due to rounding errors, the efficiency may not be exactly 100%.)
2. Set the Coeff. of friction (μ) to 0.20 and redo the experiment. Calculate the efficiency by dividing the work done on the object by the total energy (potential and kinetic) that the object gained.
3. How do you think angle and weight affect efficiency? Use the Gizmo to find out.
6. Advanced challenge: Show mathematically that the efficiency of an inclined plane is equal to 1 − (Ff / Fa), where Ff is the force of friction and Fa is the applied force.