The purpose of this lab is to determine the coefficient of kinetic friction between a block of known mass and a table surface. We will determine this using a three mass pulley system and the Work-Energy theorem.
Wood block (We will calculate the coefficient of kinetic friction of this)
Various weights
String
Cart with virtually frictionless wheels
Table
Clamps
PASCO super pulley
PASCO Capstone software
We will use a three mass system, with one of the masses on a frictionless pulley. We will keep the mass hanging off the table (see diagram) constant, while varying the mass on the wood block. We will keep the total mass of the system the same throughout the experiment, by changing \(m_{1}\) when we change \(m_{2}\).
We can use the Work-Energy theorem to create a linear model.
\[ W_{i} = F_{i}\cdot d_{i} \\ \] \[ W_{i} = F_{i}d_{i}cos(\theta_{F,d}^{\circ}) \] \[ \Sigma W = \Delta KE \\ \] \[ \Sigma W = W_{m_{1}g} + W_{F_{n1}} + W_{T_{1,2}} + W_{m_{2}g} + W_{F_{n2}} + W_{T_{2,1}} + W_{T_{2,3}} + W_{friction_{m_{2}}} + W_{m_{3}g} + W_{T_{3,2}}\\ \]
Gravitational and normal force on masses 1 and 2 are perpendicular to the objects’ displacements. \[ cos(90^{\circ}) = 0 \] Therefore, no work is done by these forces.
The tension forces between masses 1 and 2 and between masses 2 and 3 form action-reaction pairs: \[ W_{T_{1,2}} + W_{T_{2,1}} = 0\\ \] \[ W_{T_{2,3}} + W_{T_{3,2}} = 0\\ \] \[ \Sigma W = W_{friction_{m_{2}}} + W{m_{3}g}\\ \] \[ \Delta KE = F_{friction_{m_{2}}} \cdot x \cdot cos(180^{\circ}) + m_{3} \cdot g \cdot x \cdot cos(0^{\circ})\\ \] \[ \frac{1}{2} m_{total} v^{2} = -m_{2} g \mu_{k} x + m_{3} g x \]
Using this linearized theoretical model, we will graph \(\frac{1}{2} m_{total} v^{2}\) by \(-m_{2} g x\). We will keep \(m_{3}\) constant, while varying \(m_{2}\) by adding and removing various weights. The smart pulley will be used with the Capstone software to determine the velocity of the system.
If we run a linear regression on our experimental data, our mathematical model tells us that the slope of this line will be the coefficient of kinetic friction between the wood block and the desk.
We can use the same linear model to find the coefficient of static friction. We want to find the smallest \(m_{2}\) possible such that the static friction will prevent the system from smooth motion.
\[
v_{f} = 0
\]
\[
KE = \frac{1}{2} m_{total} v^{2} = 0
\]
\[
0 = -m_{2} g \mu_{s} x + m_{3} g x
\]
\[
m_{2}g \mu_{s} x = m_{3} g x
\]
\[
\mu_{s} = \frac{m_{3}}{m_{2}}
\]