1. 0 = FT + 55.1 - f 0 = 2 - FT 2N = FT 0 = 2 + 55.1 - f f = 57.1 N1)
For ANY object on an incline, Fg can be broken into two components:
F|| (Fg sinӨ) ▬ always parallel to and down the incline
FN (Fg cosӨ) ▬ always perpendicular to the incline FN
Given:
m1 = 15 kg Fg1 = 147 N
FII = 55.1 N FN = N
Fg2 = 2 N
a = 0 m/s2
FII
Fg1 FT f
F = FT + FII - f
F = m1a
m1a = FT + FII - f FT 2N
Fg2
F = Fg2 + FT
F = m2a
m2a = Fg2 - FT
0 = FT f
0 = 2 - FT
2N = FT
fs ≤ μsFN
(2) Is the same as (1) but easier do (2)
0 = f
57.1 ≤ μs(136.3)
0.4 ≤ μs
f = 57.1 N
m = 3 ⅓ kg

2. Fg FTy + FTy = Fg FTx = FTx FT(cos 50) + FT (cos 50) = 1183) FT x FT
100° x FT FT FT y y
50° FT
50° Fg
FTy + FTy = Fg
FT(cos 50) + FT (cos 50) = 118
2FT(cos 50) = 118
FT(cos 50) = 59
FTx = FTx
FT(sin 50) = FT (sin 50)
FT = FT
THANKS!
FT = 91.8 N

3. FTy = Fg FT(cos θ) = Fg 311(cos 40) = Fg FTx = F 311(sin θ) = 2004) FT θ
FTy θ F FT Fg
FTx
FTy = Fg
FT(cos θ) = Fg
311(cos 40) = Fg
FTx = F
311(sin θ) = 200
θ = 40°
Fg = 238 N

4. THESE TWO EQUATIONS AND LOTS OF ALGEBRA WILL GET YOU THE ANSWERS!5)
T1x = T2x
T1y + T2y = Fg
T1(cos α) = T2 (cos β)
T1(sin α) + T2 (sin β) = Fg
THESE TWO EQUATIONS AND LOTS OF ALGEBRA WILL GET YOU THE ANSWERS!
FT1 6)
a) T2 = 27.1 N; T1 = 33.3 N
b) α = 53.5° ; m = 2.0 kg
c) β = 69.9° ; T1 = 32.9 N; T2 = 54.8 N
d) m = 4.3 kg; T2 = 35.5 N
FT2
FT3 Fg
FT1x = FT3x
FT1y + FT2 + FT3y = Fg 7)
150N
106 N
decreases
FT1 = FT3 = 34.9 N; FT2 = 32.4 N