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phy131studiof15:lectures:m2f15info [2015/11/03 15:52] mdawber [Question 2 Solutions (23.7/35)] |
phy131studiof15:lectures:m2f15info [2015/11/03 23:14] (current) mdawber [Question 2 Solutions (23.7/35)] |
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====== Midterm 2 Information ====== | ====== Midterm 2 Information ====== | ||

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- | ===== Question 2 Solutions (23.7/35) ===== | + | ===== Question 2 Solutions (Average: 23.7/35) ===== |

I want to raise a $0.5\,\mathrm{kg}$ object from the ground by pulling on a rope that passes over a solid uniform pulley which has mass $300\,\mathrm{g}$ and radius $15\,\mathrm{cm}$. The bearing on which the pulley turns can be treated as frictionless. The 0.5 kg object is initially at rest and I would like to lift it at a constant speed of $2\,\mathrm{ms^{-1}}$. | I want to raise a $0.5\,\mathrm{kg}$ object from the ground by pulling on a rope that passes over a solid uniform pulley which has mass $300\,\mathrm{g}$ and radius $15\,\mathrm{cm}$. The bearing on which the pulley turns can be treated as frictionless. The 0.5 kg object is initially at rest and I would like to lift it at a constant speed of $2\,\mathrm{ms^{-1}}$. | ||

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$L=I\omega=\frac{1}{2}mr^{2}\omega=\frac{1}{2}\times0.3\times(0.15)^{2}\times13.33=.045\,\mathrm{kgm^{2}s^{-1}}$ | $L=I\omega=\frac{1}{2}mr^{2}\omega=\frac{1}{2}\times0.3\times(0.15)^{2}\times13.33=.045\,\mathrm{kgm^{2}s^{-1}}$ | ||

- | Out of page from right hand grip rule. | + | Direction is out of page from right hand grip rule. |

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+ | ===== Question 3 (Average: 25/35) ===== | ||

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+ | A uniform beam of mass $7\,\mathrm{kg}$ and length $80\,\mathrm{cm}$ is suspended from a rope, which is attached 20cm from the end of the beam and subject to a tension $T$. The rope makes an angle of $\theta$ with the horizontal. At its other end the beam is supported only by the normal force and the frictional force exerted by a wall (i.e. it is \textbf{not} attached to the wall by a hinge or nails or any other fastener). The beam makes a right angle with the wall. | ||

+ | | ||

+ | A. (5 points) Add arrows to the diagram showing the locations and directions of the forces which act upon the beam. Label the arrows with an appropriate descriptive variable (ie. $F_{Fr}$, $F_{N}$, $mg$, $F_{T}$) | ||

+ | | ||

+ | B. (10 points) Write equations for the sum of the horizontal forces acting on the beam, the sum of the vertical forces acting on the beam and the sum of the torques acting on the beam (calculate the torques around an axis of your choosing). (Note: Please do not just write zero three times....) | ||

+ | | ||

+ | $\Sigma F_{x}=F_{N}-F_{T}\cos\theta=0$ | ||

+ | | ||

+ | $\Sigma F_{y}=F_{Fr}+F_{T}\sin\theta-mg=0$ | ||

+ | | ||

+ | Torques calculated around contact point with wall | ||

+ | | ||

+ | $\Sigma \tau=F_{T}\sin\theta\times0.6-mg\times0.4=0$ | ||

+ | | ||

+ | | ||

+ | C. (5 points) Find the tension in the rope when $\theta$ is $10^{o}$. | ||

+ | | ||

+ | Using the torque equation | ||

+ | | ||

+ | $F_{T}$=\frac{7\times9.8\times0.4}{\sin10^{\circ}\times0.6}=263\,\mathrm{N}$ | ||

+ | | ||

+ | D. (5 points) Find the magnitude of the normal force exerted by the wall when $\theta$ is $10^{o}$. | ||

+ | | ||

+ | From the equation for the horizontal forces | ||

+ | | ||

+ | $F_{N}=F_{T}\cos10^{\circ}=263\cos10^{\circ}=259\,\mathrm{N}$ | ||

+ | | ||

+ | E. (10 points) If the coefficient of static friction of the wall is $\mu=0.2$, find the maximum value that $\theta$ can take for the beam to be in static equilibrium. (Hint: Look for a way to combine the 3 equations you wrote for part (b) so that they only contain the force due to tension and the angle $\theta$). | ||

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+ | At the maximum $\theta$, $F_{Fr}=\mu F_{N}$ so the vertical force equation is | ||

+ | | ||

+ | | ||

+ | $\mu F_{N}+F_{T}\sin\theta-mg=0$ | ||

+ | | ||

+ | Using the horizontal force equation to replace $F_{N}$ with $F_{T}\cos\theta$ | ||

+ | | ||

+ | $\mu F_{T}\cos\theta+F_{T}\sin\theta-mg=0$ | ||

+ | | ||

+ | Now I want to get rid of $mg$ which I can do using the torque equation | ||

+ | | ||

+ | $mg=F_{T}\sin\theta\frac{0.6}{0.4}=_{T}\sin\theta\frac{3}{2}$ | ||

+ | | ||

+ | Substituting this in, I get | ||

+ | | ||

+ | $\mu F_{T}\cos\theta-\frac{1}{2}F_{T}\sin\theta=0$ | ||

+ | | ||

+ | $F_{T}\sin\theta=2\mu F_{T}\cos\theta$ | ||

+ | | ||

+ | $\frac{\sin\theta}{\cos\theta}=2\mu$ | ||

+ | | ||

+ | $\tan\theta=0.4$ | ||

+ | $\theta=21.8^{\circ}$ | ||