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 downbackward 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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View Full DocumentΘ = 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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 Summer '11
 Walsh
 Force, Mass, Velocity, Vf²= Vi²+2g∆, opp/adj= Tan, Yf= Yi+

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