# s4a11 - Homework Assignment#4 Solutions Physics E-1a Fall...

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Physics E-1a Homework Assignment #4 Solutions Fall 2011 1. (a) (b) The velocity is constant so we know that the acceleration is zero a = 0 F y = F N ! " F push sin # ! F gravity = ma = 0 Fx = ! F cos "# F fr = = 0 we can find F push from this second equation. F = F cos ! or F = 60N cos 35 0 ( ) = 73.2N 2. F y = F ! N + F pull sin " # F gr = 0 and = F ! " F pull cos # = 0 \$ F = F cos # We also know that F = μ F N ! F N = F μ Now substitute this expression for F N into the F y ! equation. Which gives us F fr μ + F pull sin ! " F gr = 0 then substitute for F fr F = F cos ! F cos ! μ + F sin ! = F gr " F = F gr cos ! μ + sin ! # \$ % & ' F = mg cos 35 0 ( ) μ + sin 35 0 ( ) ! " # \$ % & We can determine μ using information from problem 1. We know that F = μ F N = 60N where F N = F sin ! + F gr plugging in numbers we find that F N = 73.2N ( ) sin 35 0 ( ) + 28kg ( ) 9.8m s 2 ( ) = 42.0N + 274.4N = 316N

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2 then, μ = F fr F N = 60 N 316 N = 0.19 Finally solve for F pull F pull = 274.4N 4.31 + 0.57 = 56.2N Pulling the cart requires less force. 3. (a) The distance the wheel travels in one revolution is the circumference of the outer edge of the wheel. C = ! D = ! 0.850m ( ) = 2.67m , if the plane is travelling at 60m/s then the wheel is revolving at 60m s ! 1rev 2.67m ! 60 min = 1348rpm rev " # \$ % (b) a centripetal = v 2 R = 60m s ( ) 2 0.850m 2 = 8471m s 2 (c) F centripetal = ma centripetal = 10 ! 15 kg ( ) 2 ( ) = 8.47 " 10 ! 12 N (d)The weight of the bacterium is mg = 10 ! kg ( ) 9.8m s 2 ( ) = 9.8 " 10 ! N The ratio of the force to the weight of the bacterium is 8.47 ! 10 " 9.8 ! 10 " = 864 1 4 . (a) The horizontal component of the normal force F N supplies the centripetal force. F
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s4a11 - Homework Assignment#4 Solutions Physics E-1a Fall...

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