{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# HW3 - Stage 0 Solutions 5.1 Model We can assume that the...

This preview shows pages 1–6. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Stage 0 Solutions: 5.1. Model: We can assume that the ring is a single massless particle in static equilibrium. Visualize: Solve: Written in component form, Newton’s first law is ( ) net 1 2 3 0 N x x x x x F F T T T = Σ = + + = ( ) net 1 2 3 0 N y y y y y F F T T T = Σ = + + = Evaluating the components of the force vectors from the free-body diagram: 1 x T = − 1 T T 2 x = 0 N 3 3 cos30 x T T = ° T 1 y = 0 N 2 y T T 2 = 3 3 sin30 y T T = − ° Using Newton’s first law: 1 3 cos30 0 N T T + ° = 2 3 sin30 0 N T T ° = Rearranging: ( )( ) 1 3 cos30 100 N 0.8666 86.7 N T T = ° = = ( )( ) 2 3 sin30 100 N 0.5 50.0 N T T = ° = = Assess: Since acts closer to the x -axis than to the y -axis, it makes sense that . 3 T r 1 2 T T > 5.5. Visualize: Please refer to the Figure Ex5.5. Solve: Applying Newton’s second law to the diagram on the left, ( ) net 2 4 N 2 N 1.0 m/s 2 kg x x F a m = = = ( ) net 2 3 N 3 N 0 m/s 2 kg y y F a m = = = For the diagram on the right: ( ) net 2 4 N 2 N 1.0 m/s 2 kg x x F a m = = = ( ) net 2 3 N 1 N 2 N 0 m/s 2 kg y y F a m = = =