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PHY 213 Homework for chapter 12 24) Redraw the diagram showing the woman on the right side of N2. The beam is not going to tip over with the woman as shown. a) Label the 4 forces acting on the beam b) Write down the EF(x,y) equations. c) When the beam starts to tip over, what is N1? d) Using this, find N2 from your Fy equation. e) With the pivot at the right fulcrum, write down the Et equation. Start with ET = ±TN1 ±TN2 ±Tgw ±Tgb = 0 %3D T = rFsine but 0 = 90 for each of these forces, so: t = rF %3D For this equation, measure rw from the right fulcrum not the left (as shown in the diagram). For the beam to tip, which side of the right fulcrum must she be on? Note: the weight of the beam acts through the center of mass. So what is the distance of the beam's center of mass from the right fulcrum? Now you should be able to get the equation for the Et.

PHY 213 Homework for chapter 12 24) Redraw the diagram showing the woman on the right side of N2. The beam is not going to tip over with the woman as shown. a) Label the 4 forces acting on the beam b) Write down the EF(x,y) equations. c) When the beam starts to tip over, what is N1? d) Using this, find N2 from your Fy equation. e) With the pivot at the right fulcrum, write down the Et equation. Start with ET = ±TN1 ±TN2 ±Tgw ±Tgb = 0 %3D T = rFsine but 0 = 90 for each of these forces, so: t = rF %3D For this equation, measure rw from the right fulcrum not the left (as shown in the diagram). For the beam to tip, which side of the right fulcrum must she be on? Note: the weight of the beam acts through the center of mass. So what is the distance of the beam's center of mass from the right fulcrum? Now you should be able to get the equation for the Et.

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Knowns: M= 90.0kg L = 6,00 meteis 4.00 metes = Normal Force n2= Normal 55.0 Icg %3D
Transcribed Image Text:Knowns: M= 90.0kg L = 6,00 meteis 4.00 metes = Normal Force n2= Normal 55.0 Icg %3D
PHY 213 Homework for chapter 12 24) Redraw the diagram showing the woman on the right side of N2. The beam is not going to tip over with the woman as shown. a) Label the 4 forces acting on the beam b) Write down the EF(x,y) equations. c) When the beam starts to tip over, what is N1? d) Using this, find N2 from your Fy equation. e) With the pivot at the right fulcrum, write down the Et equation. Start with Et = ±TN1 ±TN2 ±Tgw ±Tgb = 0 T = rFsine but 0 = 90 for each of these forces, so: t = rF For this equation, measure rw from the right fulcrum not the left (as shown in the diagram). For the beam to tip, which side of the right fulcrum must she be on? Note: the weight of the beam acts through the center of mass. So what is the distance of the beam's center of mass from the right fulcrum? Now you should be able to get the equation for the Et. f) Solve the t equation for rw. This is the distance from the right fulcrum. g) Now using the pivot at the left fulcrum, get the new equation +TN1 ±TN2 ±Tgw ±Tgb = 0. The forces will be the same, but for Et = now you will have different distances (r) for each force. For this equation, measure rw (=x) from the left fulcrum as shown in the diagram. h) Solve this equation for rw and compare it to your answer from f. At first, the answers might not seem the same (they should be). but remember how rw is measured for each part and then thev should be the same. Ok. I'll give you these so you can check. For part f) x = 1.64 m., and for part h) x = 5.64. The left and right fulcrums are 4 meters
Transcribed Image Text:PHY 213 Homework for chapter 12 24) Redraw the diagram showing the woman on the right side of N2. The beam is not going to tip over with the woman as shown. a) Label the 4 forces acting on the beam b) Write down the EF(x,y) equations. c) When the beam starts to tip over, what is N1? d) Using this, find N2 from your Fy equation. e) With the pivot at the right fulcrum, write down the Et equation. Start with Et = ±TN1 ±TN2 ±Tgw ±Tgb = 0 T = rFsine but 0 = 90 for each of these forces, so: t = rF For this equation, measure rw from the right fulcrum not the left (as shown in the diagram). For the beam to tip, which side of the right fulcrum must she be on? Note: the weight of the beam acts through the center of mass. So what is the distance of the beam's center of mass from the right fulcrum? Now you should be able to get the equation for the Et. f) Solve the t equation for rw. This is the distance from the right fulcrum. g) Now using the pivot at the left fulcrum, get the new equation +TN1 ±TN2 ±Tgw ±Tgb = 0. The forces will be the same, but for Et = now you will have different distances (r) for each force. For this equation, measure rw (=x) from the left fulcrum as shown in the diagram. h) Solve this equation for rw and compare it to your answer from f. At first, the answers might not seem the same (they should be). but remember how rw is measured for each part and then thev should be the same. Ok. I'll give you these so you can check. For part f) x = 1.64 m., and for part h) x = 5.64. The left and right fulcrums are 4 meters
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