1. I can solve for a bunch more stuff Fg= 400 N θ = 40° Fg = mg
Given:
I can solve for a bunch more stuff
Fg= 400 N
θ = 40°
Fg = mg
400 = (m) (9.8)  m = 40.8 kg
μs = 0.45
μk = 0.35
F|| = Fg sinӨ
fs ≤ μsFN
fs ≤ (.45)(306)
fs ≤ 138 N
fk =μkFN
fk = (.35)(306)
fk = 107 N
F|| = (400) sin(40º) = 257 N
FN = Fg cosӨ
For which questions are
these values valid?
FN = (400) cos(40º) = 306 N
1 through 5

2. FT = 119 N 1. Draw the free body diagram: a = 0 F = ma FN FT
F = FT + fs – F||
ma = FT + fs – F|| FN FT
F|| fs Fg
40.8(0) = FT – 257
FT = 119 N

3. FT = 364 N 2. Draw the free body diagram: a = 0 F = ma FN FT
F = FT – fk – F||
ma = FT – fk – F|| FN FT
F|| fk Fg
40.8(0) = FT – 107 – 257
FT = 364 N

4. FT = 150 N 3. Draw the free body diagram: a = 0 F = ma FN FT
F = F|| – FT – fk
ma = F|| – FT – fk FN FT
F|| fk Fg
40.8(0) = 257 – FT – 107
FT = 150 N

5. FT = 433 N 4. Draw the free body diagram: a  F = ma FN FT
F = FT – F|| – fk
ma = FT – F|| – fk FN FT
F|| fk Fg
40.8(1.7) = FT – 257 – 107
FT = 433 N

6. FT = 81 N 5. Draw the free body diagram:  a F = ma FN FT
F = F|| – FT – fk
ma = F|| – FT – fk FN FT
F|| fk Fg
40.8(1.7) = 257 – FT – 107
FT = 81 N

7. 6/7. in # 6  fk = 0 N in # 7  fk = μkFN = 31.3 N
Draw the free body diagrams:
F|| = (78.4) sin(37º)
= 47.2 N a FT FN FT a
FN = (78.4) cos(37°)
= 62.6 N m2
F|| m1 fk
Fg2
Fg1
F = m2a
F = Fg2 - FT
m2a = Fg2 - FT
F = m1a
F = FT - F|| - fk
m1a = FT - F|| - fk
in # 6  fk = 0 N
in # 7 
fk = μkFN = 31.3 N
6) a = 3.2 m/s2; FT = 73 N
7) a = 1.5 m/s2; FT = 91 N

8. 8.
Draw the free body diagrams:
F|| = (44.1) sin(37º)
= 26.5 N a FT FN FT a
FN = (44.1) cos(37°)
= 35.3 N m2
F|| m1 fk
Fg2
Fg1
fk = μkFN = 12.4 N
F = m2a
F = Fg2 - FT
m2a = Fg2 - FT
F = m1a
F = FT - F|| - fk
m1a = FT - F|| - fk
4m2 = 9.8m2 – 56.9
4.5(4) = FT – 26.5 – 12.4
FT = 56.9 N
m2 = 9.8 kg