Need a Double Check / Verification

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SUMMARY

The discussion focuses on calculating the tension in a cable and the resulting acceleration between two masses, specifically a 2 kg mass (m1) and a 5 kg mass (m2) with a coefficient of friction (μ) of 0.15. The calculations yield an acceleration of 3.85 m/s² and a tension (FT) of 11.9 N in the cable. The equations used include the force balance equations for both masses, incorporating gravitational force and frictional force. A critical error in the calculations was identified, emphasizing the importance of correctly setting equations equal to each other.

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Inferior Mind
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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
Physics Question 3 U1-C.gif


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
 
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Inferior Mind said:
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
View attachment 55355

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

Look Good !
 
No, it isn't smart guy.

5a + 7.35 =/= 5a - 7.35
 

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