Place a coin on the cover of either a binder or your textbook. Slowly raise the cover of the textbook, while observing the motionless coin. Eventually, you will reach an angle where the coin breaks free and starts sliding down the slope of the binder. This is the phenomenon we will be analyzing today.

Notable recent UC coin (cupro-nickle) masses: M_dime = 2.264g; M_penny = 3.110g; M_nickel = 4.999g; M_quarter = 5.669

Q1: Observe the coin on the binder. Before the coin slides, what kind of friction holds it in place? Sketch what a microscopic veiw of the coin and binder's interacting surfaces might look like. Continue tilting the binder until the coin starts sliding. What is the name of this new type of friction after the coin is moving and how is it different? How would your sketch change?

Static friction is what holds it in place initially. Once the coin is moving, kinetic friction is in effect with relative motion at the interface of the surfaces in contact. The sketch would change in regards to the normal force and weight are no longer in a straight line.

tan q = o/a = 66 cm/ 174 cm

tan ^-1 (66/174) = q

20.77° = q

Q2: Try to determine the angle at which the coin exactly breaks free. Is this possible? Why or why not? This is known as the angle of Repose for an object on a slope. What is the Angle of Repose you measured for your coin and the binder or book?

It is possible because a point is reached where the object's weight overcomes the static friction.  Using simple trigonometric equations and the lengths of the resulting triangle, the angle can be determined.  20.77° = q

Q3:Draw a free body diagram for the inclined plane and the coin. Choose coordinate axes so that the normal force is +y-hat and uphill is +x-hat. Identify and label the following forces on the coin: Normal force (F_N), Friction (F_s = m_kF_N)and Weight (W = mg).

Q4: Write Newton's Second Law of Motion for any object in equilibrium. What can we mathematically state about objects in equalibrium? Is the coin in equalibrium at the Angle of Repose? Before and After? Explain.

[F_NET = ma]  The sum of all forces (F_NET) is equal to zero. The coin is not in equilibrium at the angle of repose. Motion begins as the friction is overcome by weight due to the unequal forces acting on the coin.

Q5: We will use N2x and N2y. What calculations must be done to the three forces you have drawn? Which forces do not require such calculations?

All forces need to be broken down on each axis so as to be linear. The forces on each axis are set equal and the sum is found. The forces set directly on the axis do not need manipulation.

Q6: Write component expressions for N2x and N2y. What do they mean (in words)?

S F_x = [f_s - (mg sin q )] = 0

SF_y = [F_N - (mg cos q )] =0

The sum of all the forces on the x-axis are equal to the friction minus the x-component of the weight. The sum of all the forces on the y-axis are equal to the friction minus the y-component of the weight. The weight component for x and y are negative because we chose our coordinate system to be positive for f_s on the x-axis and F_N for the y-axis.

Q7: Solve for F_N and f_s alone. What do these expressions mean (in words)?

To find F_N and f_s , we will use the mass of a quarter (5.669g) and the angle 20.77°.

f_s = mg( sinq ) = (5.669g * 1kg/1,000g)(9.8m/s²)[sin 20.77°] 

f_s = (5.669kg * 10^-3)(9.8m/s²)(0.355)

f_s = 1.97 * 10^-2 N

F_N = mg(cosq ) = (5.669g * 1kg/1,000g)(9.8m/s²)[cos 20.77°]

F_N = (5.669kg * 10^-3)(9.8m/s²)(0.935)

F_N = 5.19 * 10^{-2 N}

Once these forces are ocercome on their respective axes, then a stronger force in the opposite direction will be present. f_s is the force of static friction which will hold the quarter in place until the opposite force exceeds it. F_N is the normal force pushing the quarter up on the surface.

Q8: Calculate the division N2x/N2y. Using the identity: tan q = (sin q) / (cos q) to simplify your answer. What does this mean about the role of an object's mass with regard to the Angle of Repose and the coefficient of static friction? What are the units of static friction? Does this make sense? Why?

[f_s = mg(sin q )] / [F_N = mg(cos q )] Divide and substitute in tan q = (sin q) / (cos q)

We end up with f_s/F_N = tan q : This tells us that an object will begin to slide at the angle of repose regardless of the mass of an object or it's coefficient of static friction. The mass and coefficient of static friction are proportional and that relationship remains constant with the angle of repose. The units of static friction are Newtons. Friction is a force and force is measured in Newtons. Any two sufaces that come into contact with each other will feel a given amount of friction

Q9: What do you calculate as your coefficent of static friction for the binder and coin?

f_s = m_kF_N

m_k = f_s/F_N = (1.97 * 10^-2 N)(5.19 * 10^-2 N)

m_k = 3.80 * 10^-1