Chapter 1 and 2 HW key copy

Chapter 1 and 2 HW key copy - Chapter 1 1. W= mg= 140 *...

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Chapter 1 1. W= mg= 140 * 9.81= 1373.4 N 2. M= W/g= 400/ 9.81= 40.8 kg 3. W= W(Martin)+ W (bike) = m(martin)*g + 100N= 55*9.81+ 100= 639.6 N 4. F=μR → R= F/ μ= 200/0.45= 444.4N 5. F=μR = 0.67* 1400= 938N 6. R= 400N a. μ= Fs(befor sliding)/R = 300/ 400=0.75 b. μ=Fd(sliding)/R= 200/400=0.5 7. F=μR= μmg=0.55*80*9.81= 431.6N 8. C = 1280.6N Θ = Tan -1 opp/adj= Tan -1 1000/800= 51.3degrees (51.3degree down-backward below horizon) 9. ΣF=R+(-2000N)=0→ R= 2000N (or R = W= 2000N. ..kinda) F=μR= 0.80*2000N= 1600N 10. 200²+ 1200²=C² →C= 1216.6 N Θ = Tan -1 opp/adj= Tan -1 1200/200= 80.5 drgrees from the horizon(up- forward) 11. Covert both 1000N to horizontally…one force is 60° above horizon one is 60° under F1=F2=1000*cos60°=500N F1+F2=1000N (to the right) ΣF=Fx+ (F1+F2)=0 → Fx= 1000N (to the left) (the book tells you that thepatella is in a state of static equilibrium so that means the forces have to all cancel each other out) 12. a. Sin 78°= x /4500 → x =4401.6 N b. Cos78°= y /4500 → y =935.6 N 13. 2000N southwestly can be covert to 141.4N each sides horizontaly and vertically Horizontal: F= (-141.4N)+ 100N= -41.4N Vertical: F= 150N+ (-141.4N)= 8.6N
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Θ = Tan -1 opp/adj= Tan -1 8.6/41.4= 11.7 degrees (north of west) Fr=8.6/ Sin 11.7° =42.3 N 14. The mass of the sled and juiie is 58 kg or 569 N (W=mg) Usually we put the weight in as the normal force in our friction equation but
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Chapter 1 and 2 HW key copy - Chapter 1 1. W= mg= 140 *...

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