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Magazine Chapter 1: Q72P (page 1)
Question: You are designing a system for moving aluminum cylinders from the ground to a loading dock. You use a sturdy wooden ramp that is 6.00 m long and inclined at \(37^\circ \) above the horizontal. Each cylinder is fitted with a light, frictionless yoke through its center, and a light (but strong) rope is attached to the yoke. Each cylinder is uniform and has mass 460 kg and radius 0.300 m. The cylinders are pulled up the ramp by applying a constant force \(\vec F\) to the free end of the rope. \(\vec F\) is parallel to the surface of the ramp and exerts no torque on the cylinder. The coefficient of static friction between the ramp surface and the cylinder is 0.120. (a) What is the largest magnitude \(\vec F\) can have so that the cylinder still rolls without slipping as it moves up the ramp? (b) If the cylinder starts from rest at the bottom of the ramp and rolls without slipping as it moves up the ramp, what is the shortest time it can take the cylinder to reach the top of the ramp?
Short Answer
(a)\(F = 4090.86\;{\rm{N}}\)
Step by step solution
Given Data
\(\begin{array}{l}{\rm{Length,}}\;{\rm{s}} = 6\;{\rm{m}}\\\theta = 37^\circ \\{\rm{mass}},m = 460\;{\rm{kg}}\\{\rm{radius}},r = 0.3\;{\rm{m}}\\{\rm{friction}},\mu = 0.120\end{array}\)
Concept
The summation of all the total forces acting on the particles is known as the magnitude of force.
Step 3(a): Determine the force
The expression for rolling without slipping is
\(a = \alpha r\)
\(\begin{array}{c}Fr - mgr\;\sin \;\theta = \left( {\frac{{m{r^2}}}{2} + m{r^2}} \right)\frac{a}{r}\\F - mg\;\sin \;\theta = \left( {\frac{{3ma}}{2}} \right)\\F - \left( {460\;{\rm{kg}}} \right)\left( {10\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\sin \;37^\circ = \frac{3}{2} \times 460\;{\rm{kg}} \times {\rm{a}}\\{\rm{F - 2768}}{\rm{.34 = 690a}} \cdot \cdot \cdot \cdot \cdot \cdot \left( 1 \right)\end{array}\)
Now when the friction comes into existence then the expression is,
\(\begin{array}{c}F - mg\;\sin \;37^\circ - f = ma\\F - mg\;\sin \;37^\circ - \mu N = ma\\F - \left( {460\;{\rm{kg}}} \right)\left( {10\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\sin \;37^\circ - \left( {0.120} \right)\left( {460\;{\rm{kg}}} \right)\left( {10\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}} \right)\cos \;37^\circ = \left( {460\;{\rm{kg}}} \right)a\\F - 3209.18 = 460a \cdot \cdot \cdot \cdot \cdot \cdot \left( 2 \right)\end{array}\)
From Equation (1) and (2),
\(\begin{array}{c}\frac{{F - 2768.34}}{{F - 3209.18}} = \frac{{690a}}{{460a}}\\F - 2768.34 = 1.5\left( {F - 3209.18} \right)\\F = 4090.86\;{\rm{N}}\end{array}\)
Hence, the force is \(F = 4090.86\;{\rm{N}}\)
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