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Chapter 5: Problem 58

At the end of a factory production line, boxes start from rest and slide down a \(30^{\circ}\) ramp 5.4 m long. If the slide can take no more than \(3.3 \mathrm{s}\), what's the maximum allowed frictional coefficient?

Short Answer

Expert verified
By substituting the values into the equation from Step 5, we find the maximum allowed frictional coefficient \( \mu \) is approximately 0.227.

Step by step solution

01

Set up equation of motion

The equation of motion in the direction parallel to the incline using Newton's second law is \( m \cdot g \cdot sin(\Theta) - F_{f} = m \cdot a \). On this equation \( F_{f} \) is the frictional force and \( a \) is the acceleration.
02

Express the frictional force

The frictional force \( F_{f} \) is equal to \( \mu \cdot m \cdot g \cdot cos(\Theta) \), where \( \mu \) is the frictional coefficient.
03

Combine step 1 and 2

By substituting \( F_{f} \) from step 2 into step 1, we get \( m \cdot g \cdot sin(\Theta) - \mu \cdot m \cdot g \cdot cos(\Theta) = m \cdot a \). The mass \( m \) cancels out giving \( g \cdot sin(\Theta) - \mu \cdot g \cdot cos(\Theta) = a \).
04

Use kinematic equation

Using the kinematic equation \( d = v_{i} \cdot t + 0.5 \cdot a \cdot t^{2} \) where \( d \) is distance, \( v_{i} \) is initial velocity, \( a \) is acceleration and \( t \) is time, and noting that \( v_{i} = 0 \), we get \( a = 2 \cdot d \cdot t^{-2} \).
05

Solve for \( \mu \)

Substitute the value of \( a \) from step 4 into the equation in step 3 to solve for \( \mu \), which will look like \( g \cdot sin(\Theta) - \mu \cdot g \cdot cos(\Theta) = 2 \cdot d \cdot t^{-2}, so \mu = (g \cdot sin(\Theta) - 2 \cdot d \cdot t^{-2}) / (g \cdot cos(\Theta)) \).
06

Substitute the values

Substitute \( \Theta = 30^{\circ} \), \( d = 5.4 \, m \), \( t = 3.3 \, s \) and \( g = 9.81 \, m/s^2 \) into the equation from step 5 to get the numerical value of \( \mu \).

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