This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 3.24. Visualize: 1 1 1 1 2 1 tan 45 tan 63.4 E F θ θ   = = ° = = ° Solve : (a) Thus φ = 180 °  θ E θ F = 71.6 ° 2 2 2 2 2 From the figure, 2 and 5. Using 2 cos ( 2) ( 5) 2( 2)( 5) cos(180 71.6 ) 3.00. sin sin(180 71. Furthermore, using 5 G E F EF G φ α = = = + = + °  ° ⇒ = °  = (b) E F 6 ) 45 2.975 α ° ⇒ = ° Since 45 E θ = ° , the angle made by the vector G r with the + xaxis is ( ) 45 45 90 . G E θ α θ = + = ° + ° = ° (c) We have ( 29 ( 29 2 2 1 1 1.0, and 1.0 1.0, and 2.0 0.0, and 3.0   3.0 0.0 3.0 3.0, and tan tan 90   0.0 x y x y x y y x E E F F G G G G G θ = + = + =  = + = = ⇒ = + = = = = ° That is, the vector G r makes an angle of 90 ° with the xaxis. Assess: The graphical solution and the vector solution give the same answer within the given significance of figures. 3.33. Visualize: The coordinate system ( x,y,z ) is shown here. While + x denotes East and + y denotes North, the + zdirection is vertically up. The vectors morning S r (shortened as m S r ), afternoon S r (shortened as a S r ), and the total displacement vector total a m S S S = + r r r are also shown.are also shown....
View
Full Document
 Spring '08
 Nandi
 Physics, Force, Friction, Friction Force

Click to edit the document details