0.19 Phy1170: energy -- work  (Page 7/7)

Force parallel to a ramp

A man and a crate

A man needs to load a crate with friction-free wheels and a weight of 1000 N onto the bed of a truck that is 1 m off the ground. He has two differentramps that he can use to push the crate up the ramp and onto the truck.

One ramp is 2.5 m long and the other ramp is 5 m long.

He is able to push the crate up the ramp by pushing in a direction that is parallel to the ramp.

Questions

What is the force that he must exert parallel to each ramp to push the crate up the ramp.

Which ramp requires the least amount of work to push the crate up the ramp?

Required force

Trigonometry can be used to compute the amount of force required, parallel to the ramp, to push thecrate up the ramp as a function of the angle that the ramp makes with the horizontal. That force is equal to the product of the weight of the crate (1000N or 225 pound-force) and the sine of the angle that the ramp makes relative to the horizontal.

For example, other than the requirement to overcome inertia to get the crate moving, no force is required to push the crate if the angle is 0 degrees. At theother extreme, if the angle is 90 degrees, the man must lift the crate straight up, bearing the entire weight of the crate. For angles in betweenthese two extremes, the required force is within the capability of one man to achieve.

Answers

Ramp #1

• Truck bed height = 1.0m
• Ramp length = 2.5m
• Ramp angle = arcsin(1 / 2.5) = 23.58 degrees
• Weight of crate = 1000N
• Component of crate's weight parallel to ramp = 1000 N * sin(23.58 degrees) = 400.03 newtons. This is the force that must be exertedto push the crate up the ramp.
• Work = (400.03 newtons) * 2.5 meters = 1000.08 joules

Ramp #2

• Truck bed height = 1.0m
• Ramp length = 5m
• Ramp angle = arcsin(1 / 5) = 11.54 degrees
• Weight of crate = 1000N
• Component of crate's weight parallel to ramp = 1000 N * sin(11.54 degrees) = 200.05 newtons.This is the force that must be exerted to push the crate up the ramp.
• Work = (200.05 newtons) * 5 meters = 1000.25 joules

The effect of the ramp

The man only has to exert half as much force to push the crate up Ramp #2 as is required for Ramp #1.

However, he has to push it twice as far with Ramp #2, so the amount of work done on the crate in both cases is 1000 joules.

Like many other simple machines, the use of a ramp (inclined plane) reduces the force required to do a job but it doesn'treduce the amount of work required to do the job.

Do the calculations

I encourage you to repeat the calculations that I have presented in this lesson to confirm that you get the same results. Experiment with the scenarios, making changes, and observing the results of your changes. Make certain that you can explain why your changes behave as they do.

Resources

I will publish a module containing consolidated links to resources on my Connexions web page and will update and add to the list as additional modulesin this collection are published.

Miscellaneous

This section contains a variety of miscellaneous information.

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