In the figure below a box (total mass m_1 = 1.65 kg) and another box (total mass m_2 =3.30 kg) slide down an inclined plane while attached b a massless rod parallel to the plane. The angle if incline is \Theta = 30.0°. The coefficient of kinetic friction between m_1 and the incline is \Theta =30.0°. The coefficient of kinetic friction between \mu_1 and the incline is \mu_1 = .226; that between m_2 and the incline is \mu_2 = .113. Compute the tension in the rod and the magnitude of the common acceleration of the two boxes.

Once the force vectors are labeled and the axis components are resolved we can set up Force Systems to compute the tension and magnitude of common acceleration. The systems are as follows:

System 1a: m1, perpendicular, a = 0 

Force balance, |C_1| = |m_1gcos \theta| .

System 1b: m1, parallel, a

+|m_1gsin \theta|+|T|-|F_{f1}|=m_1a

System 2a: m2, perpendicular, a = 0

Force balance, |C_2| = |m_{2}gcos \theta|

System 2b: m2, parallel, a

+|m_2gsin \theta| - |T| - |F_{f2}|=m_2a

F_{f1} Model

|F_{f1}| = .226C_1

F_{f2} Model

|F_{f2}| = .113C_2

Since there is no acceleration in the vertical dimension we only add the horizontal systems for both masses to obtain \sum F of the system. This gives us:

gsin \theta(m_1+m_2) - F_{f1} - F_{f2} = (m_1 + m_2)a

a = g[sin 30(m_1+m_2) - (\mu m_1 + \mu m_2)cos 30

a = 3.62m/s^2

|T| = |m_1a| + |F_f|-|m_1gsin \theta |

|T| = |(1.65)(3.62)| + |.726m_1gcos30|-|(1.65)(9.8)sin30|

|T| = 1.05 N

Arc Length

In this post I want to explain how a definite integral can be used to find the arc length of a curve because I think that it is fascinating that integrals have functions other than just finding the area under curves. Let us start with a definition:

A RECTIFIABLE curve is one that has a finite arc length. A sufficient condition for the graph of a function f to be rectifiable between (a, f(a)) and (b, f(b)) is that  f' be continuous on [a, b] . Such a function is continuously differentiable on [a, b] , and its graph on the interval [a, b] is a smooth curve.

Now let’s consider a function y = f(x) that is continuously differentiable on the interval [a, b] . We can approximate the graph of f by line segments whose endpoints are determined by the partition:

a = x_{0} < x_{1} < x_{2} <....< x_{n} = b

By letting \Delta x_{i} = x_{i} - x_{i-1} and \Delta y_{i} = y_{i} - y_{i-1} , we can approximate the length of the graph by the following:

\displaystyle S \approx \sum_{i=1}^{n} \sqrt{(x_{i} - x_{i-1})^2 +(y_{i} - y_{i-1})^2}

= \displaystyle \sum_{i=1}^{n} \sqrt{(\Delta x_{i})^2 + (\Delta y_{i})^2}

= \displaystyle \sum_{i=1}^{n} \sqrt{(\Delta x_{i})^2 +(\frac{\Delta y_{i}}{\Delta x_{i}})^2(\Delta x_{i})^2}

= \displaystyle \sum_{i=1}^{n} \sqrt{1 + (\frac{\Delta y_{i}}{\Delta x_{i}})^2}(\Delta x_{i})

This approximation becomes more and more precise as ||\Delta|| \to 0 (n \to \infty) . Therefore the length of the graph is S = \displaystyle \lim_{||\Delta|| \to 0} \displaystyle \sum_{i=1}^{n} \sqrt{1 + (\frac{\Delta y_{i}}{\Delta x_{i}})^2}(\Delta x_{i})

Because f'(x) exists for each x in (x_{i-1}, x_i) , the Mean Value Theorem guarantees the existence of c_i in (x_{i-1}, x_i) such that f(x_i) - f(x_{i-1}) = f'(c_i)(x_i - x_{i-1})

\frac{\Delta y_i}{\Delta x_i} = f'(c_i)

Because f' is continuous on [a, b] , it follows that \sqrt{1 + [f'(x)]^2} is also continuous and therefore integrable on [a, b] , which implies that S = \displaystyle \lim_{||\Delta|| \to 0} \displaystyle \sum_{i=1}^{n} \sqrt{1 + [f'(c_i)]^2}(\Delta x_i)                                                                               = \int_{a}^{b} \sqrt{1 + [f'(x)]^2} dx