Solutions 7

# Solutions 7 - 5.43(a IDENTIFY Apply F = ma r r to the crate...

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5.43. (a) IDENTIFY: Apply m = F a r r to the crate. Constant v implies 0. a = Crate moving says that the friction is kinetic friction. The target variable is the magnitude of the force applied by the woman. SET UP: The free-body diagram for the crate is sketched in Figure 5.43. EXECUTE: y y Fm a = sin 0 nm gF θ −= sin = + kk k k sin fn m g F µµ µ == + Figure 5.43 x x a = k cos 0 Ff cos sin 0 g F θµ −− = (cos sin ) g θµ θ µ k k cos sin mg F = (b) IDENTIFY and SET UP: “start the crate moving” means the same force diagram as in part (a), except that k is replaced by s . Thus s s . cos sin mg F = EXECUTE: F →∞ if s cos sin 0. θµ θ This gives s cos 1 . sin tan EVALUATE: F r has a downward component so . g > If 0 = (woman pushes horizontally), g = and . m g 5.45. IDENTIFY: Apply m F =a r r to each block. SET UP: For block B use coordinates parallel and perpendicular to the incline. Since they are connected by ropes, blocks A and B also move with constant speed. EXECUTE: (a) The free-body diagrams are sketched in Figure 5.45.

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Solutions 7 - 5.43(a IDENTIFY Apply F = ma r r to the crate...

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