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 phy131studiof15:lectures:m2p2sol [2015/10/26 09:22]mdawber [Question 2 Solution] phy131studiof15:lectures:m2p2sol [2015/10/26 09:35]mdawber [Question 3 Solution] Both sides previous revision Previous revision 2015/10/26 09:38 mdawber 2015/10/26 09:35 mdawber [Question 3 Solution] 2015/10/26 09:22 mdawber [Question 2 Solution] 2015/10/26 09:18 mdawber [Question 1 Solution] 2015/10/22 14:53 mdawber [Question 2 Solution] 2015/10/22 14:53 mdawber [Question 2 Solution] 2015/10/20 11:53 mdawber [Question 2 Solution] 2015/10/20 11:51 mdawber 2015/10/20 11:51 mdawber [Midterm 2 Q2 Solution (Ave Score: 29.1/35)] 2015/10/20 11:49 mdawber [Question 1 Solutions] 2015/10/20 11:49 mdawber [Question 1 Solutions] 2015/10/20 11:48 mdawber created 2015/10/26 09:38 mdawber 2015/10/26 09:35 mdawber [Question 3 Solution] 2015/10/26 09:22 mdawber [Question 2 Solution] 2015/10/26 09:18 mdawber [Question 1 Solution] 2015/10/22 14:53 mdawber [Question 2 Solution] 2015/10/22 14:53 mdawber [Question 2 Solution] 2015/10/20 11:53 mdawber [Question 2 Solution] 2015/10/20 11:51 mdawber 2015/10/20 11:51 mdawber [Midterm 2 Q2 Solution (Ave Score: 29.1/35)] 2015/10/20 11:49 mdawber [Question 1 Solutions] 2015/10/20 11:49 mdawber [Question 1 Solutions] 2015/10/20 11:48 mdawber created Last revision Both sides next revision Line 100: Line 100: {{m2fig2f11.png}} {{m2fig2f11.png}} - A. Sum of torques on beam around hinge + A Jack-O'​-Lantern of mass 4kg is to be suspended as shown in the diagram using a hinged uniform beam of mass 2kg and length 0.8m and a massless string. The  beam should be level and the string at 30º to the horizontal. + + (a) (10 points) If the maximum tension the string can support without breaking is 50N, what is the furthest distance from the hinge, $x$, that I can hang the Jack-O'​-Lantern. + + Sum of torques on beam around hinge $0.8\mathrm{m}\times T\sin30^{o}-x\times 4\mathrm{kg}\times g-0.4\mathrm{m}\times 2\mathrm{kg}\times g=0$ $0.8\mathrm{m}\times T\sin30^{o}-x\times 4\mathrm{kg}\times g-0.4\mathrm{m}\times 2\mathrm{kg}\times g=0$ Line 106: Line 110: $x=\frac{0.8\mathrm{m}\times 50\mathrm{N}\sin30^{0}-0.4\mathrm{m}\times 2\mathrm{kg}\times g}{4\mathrm{kg}\times g}=0.31\mathrm{m}$ $x=\frac{0.8\mathrm{m}\times 50\mathrm{N}\sin30^{0}-0.4\mathrm{m}\times 2\mathrm{kg}\times g}{4\mathrm{kg}\times g}=0.31\mathrm{m}$ - B. Sum of horizontal forces on beam + (b) (10 points) If I hang the Jack-O'​-Lantern at the distance found in part (a), (ie. when the tension in the string is 50N), what is the magnitude of the net force on the hinge? What is the direction of the force (give your answer in terms of the angle $\theta$ from the $y$ axis shown in the diagram)? + + Sum of horizontal forces on beam $F_{HH}-T\cos30^{o}=0$ $F_{HH}-T\cos30^{o}=0$ Line 124: Line 130: But the force **on** the hinge is directed in the opposite direction $180^{o}-52^{o}=128^{o}$ But the force **on** the hinge is directed in the opposite direction $180^{o}-52^{o}=128^{o}$ - C. Sum of torques on beam around hinge + + (c) (10 points) If I would like to hang the Jack-O'​-Lantern at the far end of the beam from the hinge using the same string as was used in parts (a) and (b), what is the minimum angle the string should make with the horizontal instead of the $30^{o}$ angle it makes in parts (a) and (b). + + Sum of torques on beam around hinge $0.8\mathrm{m}\times T\sin\phi-0.8\mathrm{m}\times 4\mathrm{kg}\times g-0.4\mathrm{m}\times 2\mathrm{kg}\times g=0$ $0.8\mathrm{m}\times T\sin\phi-0.8\mathrm{m}\times 4\mathrm{kg}\times g-0.4\mathrm{m}\times 2\mathrm{kg}\times g=0$ 