Inferior Mind
- 14
- 0
In the following diagram there is a coefficient of friction, μ, of 0.15 between the 5.0 kg mass and the surface. Calculate the tension in the cable connecting the two masses and the resulting acceleration.
FBD
m1= 2 kg
m2= 5 kg
μ= .15
g= 9.8 m/s2
Equation 1 ~
Fg - FT = FN
2kg × 9.8m/s2 - 2kg(a) = FT
Equation 2 ~
Fg = FN = 5kg (9.8 m/s2)
~ μFN = .15(49) = 7.35 N
FT - FF = ma
FT = 5kg(a) + 7.35
~Set Equations Equal to Each Other~
19.6 - 2a = 5a - 7.35
26.95 = 7a
a = 3.85 m/s2
Sub into Eq to get FT on String
2 (9.8) - 2(3.85) = FT
FT = 11.9 N
FBD
m1= 2 kg
m2= 5 kg
μ= .15
g= 9.8 m/s2
Equation 1 ~
Fg - FT = FN
2kg × 9.8m/s2 - 2kg(a) = FT
Equation 2 ~
Fg = FN = 5kg (9.8 m/s2)
~ μFN = .15(49) = 7.35 N
FT - FF = ma
FT = 5kg(a) + 7.35
~Set Equations Equal to Each Other~
19.6 - 2a = 5a - 7.35
26.95 = 7a
a = 3.85 m/s2
Sub into Eq to get FT on String
2 (9.8) - 2(3.85) = FT
FT = 11.9 N