This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: PHY2048 Fall 2009 Homework Set 5 Solutions University of Florida Page 1 of 6 Department of Physics PHY 2048 Homework Set 5 Solutions (Chapter 6) Problem 65*: Here we sum all the vector forces on the crate and set the net force equal to the mass times the acceleration a m F F N i i net r r r = = ∑ = 1 (eq. 1) . Looking at the x component of (eq. 1) yields Ma F s − = − , where M is the mass of the crate and where I set a x = a ( i.e. deceleration). Looking at the y component of (eq. 1) yields = − Mg F N and hence Mg F N = since a y = 0. We know that (F s ) max = μ s F N and hence 2 2 max max / 45 . 2 ) / 8 . 9 )( 25 . ( ) ( s m s m g M Mg M F M F a s s N s s = = = = = = μ μ μ . Using the kinematic equations of motion we have t a v t v x x max ) ( ) ( − = 2 max 2 1 ) ( ) ( t a t v t x x − = Where I let x = 0 at t = 0 and where s m km m s hr hr km v x / 33 . 13 10 600 , 3 1 / 48 ) ( 3 = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ × ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ × = . Let t s be the time the train stops. Hence, s x s x t a v t v max ) ( ) ( − = = and thus max ) ( a v t x s = . Let d be the distance travelled in the time t s . Hence, m s m s m a v t a t v t x d x s s x s 3 . 36 ) / 45 . 2 ( 2 ) / 33 . 13 ( 2 ) ( ) ( ) ( 2 2 max 2 2 max 2 1 = = = − = = Problem 614*: Here we sum all the vector forces on the block and set the net force equal to the mass times the acceleration a m F F N i i net r r r = = ∑ = 1 (eq. 1) . First lets calculate the minimum static coeffiecient of friction such that the block does not move (i.e. a x = 0). Looking at the xcomponent of (eq. 1) yields cos = = − x s Ma F F θ , where M is the mass of the block. Looking at the ycomponent of (eq. 1) yields sin = − + Mg F F N θ since a y = 0. We know that F s = μ s F N and hence ) sin ( cos = − − θ μ θ F Mg F s (eq. 1) and solving for μ s gives F xaxis yaxis F N F g θ F k F s xaxis yaxis F N F g crate V PHY2048 Fall 2009 Homework Set 5 Solutions University of Florida Page 2 of 6 Department of Physics 5668 . ) 20 sin( 2 ) 20 cos( sin 5 . cos 5 . sin cos ) ( min = − = − = − = o o θ θ θ θ μ Mg Mg Mg F Mg F s . (a) we see that μ s = 0.6 > 0.5668 and hence a x = 0....
View
Full Document
 Fall '08
 Field
 Physics, Acceleration, Force, Mass, Work, Trigraph, EQ, Department of Physics

Click to edit the document details