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Per Lindberg · Answer 1 · 2023-05-21T01:52:28+0000

We are asked to determine the coefficient of friction in the sysmtem. First we will do a free body diagram of the first mass, like this:

Where:

$\begin{gathered} N=\text{ Normal Force} \\ m_1=\text{ mass} \\ g=\text{ acceleration of gravity} \\ T=\text{ tension} \\ F_f=\text{friction force} \end{gathered}$

Now, we add the forces in the horizontal direction, we get:

Now, to determine the friction force we need to use the following relationship:

$F_f=\mu N$

To determine the normal force we will add the forces in the vertical direction. Since there is no movement in the vertical direction this sum must add up to zero:

$\begin{gathered} N-m_1g=0 \\ N=m_1g \end{gathered}$

Now, we substitute the value of the normal force in the equation for the friction force:

$F_f=\mu m_1g$

Now, we substitute the friction force in the sum of horizontal forces:

$T-\mu m_1g=m_1a$

Now, we turn our attention to the second mass. We add the forces in the vertical direction:

Now, since the acceleration "a" and the tension "T" is the same for both masses we will solve fot "T" in the sum of forces for the second mass, like this:

Now, we substitute the value in the sum of forces of the first mass, we get:

$m_2g-m_2a-\mu m_1g=m_1a$

Now, we solve for the coefficient of friction. To do that we will subtract "m2g" to both sides:

$-m_2a-\mu m_1g=m_1a-m_2g$

Now, we add "m2a" to both sides:

$-\mu m_1g=m_1a-m_2g+m_2a$

Now, we divide both sides by "-m1g":

$\mu=(m_1a-m_2g+m_2a)/(-m_1g)$

We have gotten an expression for the coefficient of friction but we need to determine the acceleration of the system. To do that we will use the fact that the second mass moves 1 meter in 1.33 seconds. Assuming constant acceleration we can use the following equation of motion: