18-March-2015: Modeling Friction Forces

Lab 6: Modeling Friction Forces Lab

Purpose: To understand the different types of friction and to develop models that predict the coefficients of friction.

Part 1: Static Friction

Static friction describes the friction force acting between two bodies when they are not moving relative to one another.

Static Friction Force (Fs) = Coefficient of Friction (Us)* Normal Force (N)

Fs is less than or equal to Us*N

 

 

For this portion of the lab, we had to determine the coefficient of static friction between a block and the table top.

 

 

Let's say the picture above is in equilibrium. This means that normal force provided by the cup is not  enough to make the block. From this perspective, you cannot find the coefficient of static friction.

 

In order to determine the coefficient of static friction, we slowly added water to cup to the point that it would make the block move. At this point, the weight of the cup was greater than the static force, which in turn made the block move. We repeated this process four times, increasing the mass of the block each time. In doing so, we recorded the mass of the blocks and the mass of the cup with the added water for each trial.

Mass of Cup= 2.9 +/- .1g

• Trial 1: M(1 block)= 125.7 +/- .1 g , Mcup+water= 54.0 +/- .1 g
• Trial 2:  M(2 blocks)= 260.1 +/- .1 g , Mcup+water= 95.2 +/- .1 g
• Trial 3 : M(3 blocks)= 381.4 +/- .1 g , Mcup+water= 120.7 +/- .1 g
• Trial 4: M(4 blocks)= 516.4 +/- .1 g , Mcup+water= 172.8 +/- .1 g

 We then inputted these results into a graph on LoggerPro.



 

Normal Force of blocks(x) vs. Static Friction Force (y)

 

• Normal Force (x): Normal Force of Blocks = (Mass of block/1000)*9.81
• Static Friction Force (y): Weight of Cup = (Mass of cup+water/1000)*9.81

We divided the given masses by 1000 to convert grams to kilograms

 



This setup of x and y will give us the coefficient of static friction of the block. The slope of the graph equals the coefficient of static friction. In this case the slope of the graph is 0.2958.



 

Coefficient of static friction (Us) = 0.30

 

 

Part 2: Kinetic Friction

 

For this portion of the lab, we assume that kinetic friction is proportional to the normal force, and independent of the area or speed of the moving object.

Therefore: Force of kinetic friction (Ff)= Coefficient of kinetic friction (Uk)* Normal Force (N)

 

In order to find the coefficient of kinetic friction, we tied a force sensor to a block with a piece of string and dragged the block along the tabletop with constant velocity.

 







Before running the experiment, we calibrated the sensor by using a .5 kg hanging mass.

 

With the sensor calibrated, we then dragged it along the table with constant velocity and recorded the results. The force registered by the sensor is the friction force of the block. We did this step for 1,2,3, and 4 blocks.

• M1=124.9 +/- .1 g
• 
  
  M2=167.5 +/- .1 g
• 
  
  M3=149.2 +/- .1 g
• 
  
  M4=106.1 +/- .1 g

 

Measured Friction Forces



Each line represents a different trial. We then took an average of each run, to determine each block's friction force.

• Run 1 (M1): Friction Force (Ff) = .3447
• Run 2 (M1+M2): Friction Force (Ff) = .6857
• Run 3 (M1+M2+M3): Friction Force (Ff) = .9356
• Run 4 (M1+M2+M3+M4): Friction Force (Ff) = 1.198

Like in Part 1, we used the measured data to create a Friction Force vs. Normal Force graph. The slope of this graph will give the coefficient of kinetic friction for the block.

Normal Force of blocks(x) vs. Friction Force (y)

• Normal Force (x): Normal Force of Blocks = (Mass of block/1000)*9.81
• Friction Force (y): Measured Friction Force for each block (given in previous step)

Slope of graph is .2247+/- .006835. Therefore Uk = 0.2247+/- .006835

Part 3: Static Friction From A Sloped Surface

In this portion of the lab, we determined the coefficient of static friction for a block on a sloped surface. We placed a blocked on a horizontal surface and slowly raised one end of the surface, tilting it until the block started to slip. At this point, we recorded the angle of the surface at which the block began to slip. We then drew an FBD of the block on a sloped surface and solved for forces in the y and x-direction. From here, it was rather simple to solve for the coefficient of static friction.

 

First we weighed the block, m = 0.1249 kg. Then, we measured the angle at which the block began to slip, theta= 18 +/- 2 degrees. Finally, we drew an FBD of the mass and solved for forces in the y and x-direction. Here, we solved for the coefficient of static friction. Us = .33

Part 4: Kinetic Friction From Sliding A Block Down an Incline

In this part of the lab, we set up the block on an incline with a motion sensor. The block was released and the motion sensor then measured the block's acceleration. This helped us determine the coefficient of friction between the block and the incline.

 

 

First, we weighed the block. Then, we measured the angle of the incline.

• m= 168.9 +/- .1 g
• Theta= 21.3 +/- .1 degrees

Then we used the motion sensor to measure the block's acceleration.

• a=.8091 m/s^2

From here, we drew an FBD of the block and solved for the force in the y and x-direction.

First, we plugged in our known values and then simplified the equation. Finally, we solved for Uk.

• Coefficient of friction (Uk) = .30

Part 5: Predicting The Acceleration Of A Two-Mass System

In this portion of the lab, we used the calculated coefficient of friction from part 4 to derive an expression that would predict the acceleration of the system shown below.

As you can see the block is positioned on a flat, horizontal surface. The motion sensor is placed at one end of the track. In this scenario, the hanging mass will cause the block to accelerate and the motion sensor will record the acceleration. From here, we drew and FBD and solved for acceleration. If done right, the calculated acceleration and measured acceleration should match.

First, we weighed the block and the hanging mass.

• Mass of block = 168.9 +/- .1 g
• Hanging mass = 70 +/- .1 g

We then measured the acceleration of the system with the motion sensor.

• a = .2775 m/s^2

Next, we solved for acceleration with an FBD of the system.

First, we solved for the x and y forces of each block. We then set them equal to each other because the unknown Tensions (T) are the same. We also inputted Uk from the previous step. From here, we combined like terms and solved for acceleration.

• Calculated Acceleration = .79 m/s^2

To recall, our measured acceleration was 0.2775 m/s^2.

As you can see, our accelerations do not match. We suspect the error comes from our measured acceleration. The logic behind the calculated acceleration is sound. We calculated the acceleration three times and came up with the same result. The uncertainty in our measurements is small enough that we do not consider it as the source of error, since our accelerations are far apart.Therefore, we assume that the measured acceleration is incorrect.