PHYS_2014_Homework_3_soln

PHYS_2014_Homework_3 - PHYS2014 Benton Fall 2010 OSU Physics Dept Physics 2014 General Physics I Homework Assignment 3 Due Friday 17 September 2010

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PHYS2014 Fall 2010 Benton OSU Physics Dept. 1 Physics 2014: General Physics I Homework Assignment 3 Due: Friday, 17 September 2010 at the beginning of your Recitation session. 1. A 1000 kg mass is suspended from two chains as illustrated in the figure below. The chain on the left is at a 25 ° angle with respect to the ceiling, while the chain on the right is at a 45 ° angle with respect to the ceiling. Find the tension force in each of the two chains. How are the horizontal (x) components of force in the two chains related? Solution: Since the mass and chains are all at rest, there is no net acceleration, meaning that this is a case of static equilibrium and that 0 F = a . First we need to draw a free body diagram and label all the forces. Since the tension forces in the two chains are at arbitrary angles, we need to find the x and y components of the tension forces. 1000 kg 25 ° 45 ° g F a 1 T F a 2 T F a 25 ° 45 ° 25 ° 1 T F a 1 xT F a 1 yT F a 45 ° 2 T F a 2 xT F a 2 yT F a
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PHYS2014 Fall 2010 Benton OSU Physics Dept. 2 1 1 1 1 2 2 2 2 1 1 2 2 cos sin cos sin xT T yT T xT T yT T F F F F F F F F θ = = = = ( ) ( ) 1 2 1 2 1 2 1 2 1 2 1 2 2 2 2 2 2 1 2 1 2 2 1 1 2 2 1 2 1 2 1 2 2 1 2 0 0 cos cos 0 cos cos cos cos sin sin 0 cos sin sin cos cos tan sin cos tan sin 1000kg 9.8m s x xT xT y g yT yT T T T T T T T T T T T T T F F F F F F F F F F F F F mg F F F F mg F mg mg F F = = + = = + + + = = = + + = + = + = = + = ( ) 1 2 2 2 1 9452N cos45 tan 25 sin 45 cos cos45 9452N 7374N cos cos25 T T F F = + = = = r r r r r The x-components of tension in the two chains are of equal magnitude, but opposite direction. 1 2 1 2 cos cos 7374Ncos25 9452Ncos 45 6683N T T F F = = = r r 2. Can your car stop on a dime? That’s what the salesman said. If your car has a mass of 2000 kg and you are driving at a constant velocity of 120 km/hr when you slam on the breaks, what force must the breaks exert to bring the car to rest in a distance equal to 1.7 cm diameter of a dime? How many g’s is this and what effect is this force likely to have on you, the driver? Assume that the wheels don’t lock and the car doesn’t skid (admittedly not very realistic). Solution: using a kinematic equation, we can find the value of acceleration that will bring a car
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This note was uploaded on 01/10/2012 for the course PHYS 2014 taught by Professor Nandi during the Fall '08 term at Oklahoma State.

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PHYS_2014_Homework_3 - PHYS2014 Benton Fall 2010 OSU Physics Dept Physics 2014 General Physics I Homework Assignment 3 Due Friday 17 September 2010

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