Phun with Physics: UFPM Summary

Unbalanced Force Particle Model

In this unit, we summed up a lot of our previous knowledge on free body diagrams and forces and applied that knowledge as we dove deeper into the relationships between acceleration, mass, and force defined by Newton’s Second Law. Later on in the unit we also brought friction back into the picture as we refreshed our memories on how to use the formula and solve for certain variables.

Elevator Lab

One of the first experiments of this unit was performed in an elevator as we tested the weight of one subject and how it fluctuates as the elevator speeds up or slows down. Each group used a scale and measured the weight of an individual throughout different stages of the elevator’s travel. The results of my group are listed below:

Original weight: 121.4 lbs

At rest at the bottom: 121.4

Starting to go up: 134.6

Going up at constant speed: 121.6

Slowing to stop at the top: 112

Stopped at the top: 121.4

Starting to go down: 111.8

Going down at constant speed: 121.6

Slowing to stop at the bottom: 130.6

The data shows that the weight stayed constant when at rest and traveling at constant velocity. When starting to go up the weight increases, but then vastly decreases right before it comes to a stop. It is the opposite when going down, as it vastly decreases when first starting to accelerate downward, then increases when it slows to a stop. Through this experiment, we were able to learn that Fg, or force of gravity that the person exerts on the floor will always stay constant, but the force the floor exerts on the person varies which is what determines acceleration. If the elevator is accelerating upward, the Fn or normal force from the floor is greater, indicated by a longer force vector of normal force and shorter for Fg in a free body diagram. If the elevator is accelerating downward, the Fg is larger indicated by a longer vector for gravity. We also learned that accelerating upward can be viewed as a positive acceleration and accelerating downward can be viewed as a negative acceleration.

Skateboard Trials

Another introductory experiment we did to introduce us to this unit was our test runs on the skateboard. One person sat on the skateboard while another pulled them forward. We ran several trials on people of various weights to see the difference amounts of force that had to be applied to get the person to accelerate. We came to the conclusion that the greater the mass, the greater the force needed to accelerate the system, and the greater the force applied, the faster the acceleration.

Newton’s Second Law Post Lab Questions and Important Rules for the Unit

After completing both the elevator and the skateboard tests, we were able to learn how the information was related using Newton’s 2nd Law. We also familiarized ourselves with many of the important terms and methods we would be using thought the unit which will be stated below and referenced throughout the rest of the blog. Other key points from outside of the post lab questions will also be listed below.

Newton’s 2nd Law: ~therefore, the greater the force applied to the object, the faster it will accelerate, but the heavier the object is, the slower that acceleration will be

Net Force: the forces on the object that are responsible for accelerating the object, or the unbalanced force

~if a baby is sitting in a wagon and her mother is pulling the wagon (assuming there is no friction), the net force would be the amount of pull the mom applies to the wagon because the forces in the other directions are balanced

Formula for calculating acceleration: a(acceleration)=F Net/mass(kg)

~when using this formula, remember that acceleration is found in meters per second squared or m/s^2, F Net is always and ONLY the forces that are accelerating the system, and mass is always found in kilograms, so conversion between newtons and kg is often needed {(kg)(10)=N}

~F Net is a special case because the “base form” for representing it is Fn+(-Fg) which is only used when the unbalanced force is in the vertical direction. Fg is described as being negative because it is pointing downward on a free body diagram. The same would apply for friction, because it is always opposing the direction of motion, is is indicated as negative, or away from the positive, which is the direction of motion. The best way to remember this is to make a key for oneself labeling the axes with positive and negatives that correlate with he forces.

If mass doubled- acceleration would be cut in half due to inverse proportions

If force doubled- acceleration would double due to direct proportions

F Net and acceleration are always in the same direction and are directly proportional, but any forces off the axis do not count as the Net Force

weight(Newtons)=(mass)(gravitational constant)

~EX: person weighs 11kg, the gravitational constant is 10, so the weight= (11kg)(10)=110N

Formula for friction: Ff=(Mk coefficient of friction)(Fn)

UFPM Worksheet 1+2: With these 2 worksheets, we started with the basics and calculated simple net force problems where we were asked to solve for just one element of the net force such as the FN with the given information using a free body diagram. The questions below are taken straight from worksheet 1.

As pictured above, the free body diagram was created with a slanted axis because it depicts a child going down a slide, pay special attention to the way in which the forces were labeled, either positive or negative. the 55 degree angle was placed in between Fg and Fgy as normal, and 35 degrees was placed next to it so they both add to equal a 90 degree angle. In this situation, we are assuming that Fgy and Fn are equal and opposite forces, therefore, the net force on the child will be the combination of the frictional force of 160N and the force from Fgx. Fgx was found using the cosine of 35 degrees, which gave a result of 245.74 N. As stated earlier, friction is negative because it opposes the direction of motion, so a positive 245.74N Fgx added with a negative Ff of 160 N {(245.74 +(-160)} gets the net force of 85.74 Newtons which accelerates the child.

When finding the acceleration, the formula a=fnet/mass is used. We previously solved for the Fnet above, so the 85.74 N can be plugged into the equation divided by the mass of the child which is 30 kg. REMEMBER TO USE KG, NOT NEWTONS!!! Divide the two and that is the acceleration.

Above is the FBD for the grocery bag. 20 kg is converted to 200N by multiplying by 10. We do not know what the Fn force is, but we are given the acceleration, so we can use the formula to solve for FN=n. 5 is set equal to Fn-200/20. Multiply both sides by 20 and then add 200 to get the Fn of 300N, which answers the question that the groceries will not stay in the bag if the maximum capacity it can carry is 250N.

UFPM 3: In this worksheet, we brought back the formulas from the CAPM units to combine with out new study. These word problems involve collecting the data that was given and then picking the correct formula to use to solve for the necessary information. Below is a sample problem from the worksheet.

Below are the formulas we brought back from the previous unit:

As pictured, the variables given in the problem were listed out to the side along with a FBD to better illustrate the situation. The Fnet was used to solve for acceleration which was then used in another equation along with the other variables to solve for displacement.

UFPM Worksheet 4: This worksheet was quite similar to worksheet 3, but there was a lot more with correctly placing angles in the FBD.

In the diagram above, a person is shown applying a horizontal force to the box. When drawing the FBD, the angle is placed opposite to the 20 degree angle, otherwise known as vertical angles in geometry, this is indicated in purple pen.

Below are two more examples of properly placing angles in FBD’s.

UFPM Worksheet 5: In this worksheet we began to use friction and the coefficient of friction to solve problems.

Pictured above, a free body diagram is illustrated, and the coefficient of friction is given so friction can be solved for. .15 is multiplied by the Fn of 500, and Ff came out to be 75 N which was then made negative on the FBD.

Acceleration was then calculated in the same way with net force over mass.

Atwood Machines: When solving with Atwood machines, there are a few things to keep in mind.

When solving for acceleration, the net force is the tension pulling the system, or all the forces combined depending on if it is a modified Atwood machine or a regular one.

Modified: a=Fnet/mass

a= 20N (because that is the weight of the hanger)/5 kg (because this is the weight of the whole system)