Another Double Check FBD String Tension

AI Thread Summary
The discussion focuses on calculating the tension in a cable connecting two masses, m1 (5 kg) and m2 (6 kg), on frictionless surfaces at angles θ1 (60°) and θ2 (70°). The equations derived show that the tension (FT) is influenced by the gravitational force and the acceleration of the masses. After setting the equations equal, the acceleration is determined to be 1.16 m/s², leading to a final tension of 48.2 N in the cable. Concerns are raised about using the same variable names for different mass quantities, but clarification is provided that FT and acceleration can be consistently used across both masses. The calculations and results are confirmed to be correct.
Inferior Mind
Messages
14
Reaction score
0
Calculate the tension in the cable connecting the two masses. Assume all surfaces are frictionless.

FBD
Physics Question 4 U1-C.gif


m1 = 5 kg
θ1 = 60°
m2 = 6 kg
θ2 = 70°

Equation 1 ~

FT - Fg = ma

FT - mgSinθ = ma

FT = 5a + 5(9.8)Sin60

FT = 5a + 42.44

Equation 2 ~

Fg - FT = ma

mgSinθ - FT = ma

6(9.8)Sin70 - 6a = FT

FT = 55.25 - 6a

~Set Equations Equal to Each Other ~

5a + 42.44 = 55.25 - 6a

11a = 12.81

a = 1.16 m/s2

~Sub into Eq to find Force Tension on da String Son !~

5(1.16) + 5(9.8)Sin60 = FT

FT = 48.2 N
 
Physics news on Phys.org
Inferior Mind said:
Calculate the tension in the cable connecting the two masses. Assume all surfaces are frictionless.

FBD
View attachment 55356

m1 = 5 kg
θ1 = 60°
m2 = 6 kg
θ2 = 70°

Equation 1 ~

FT - Fg = ma

FT - mgSinθ = ma

FT = 5a + 5(9.8)Sin60

FT = 5a + 42.44

Equation 2 ~

Fg - FT = ma

mgSinθ - FT = ma

6(9.8)Sin70 - 6a = FT

FT = 55.25 - 6a

~Set Equations Equal to Each Other ~

5a + 42.44 = 55.25 - 6a

11a = 12.81

a = 1.16 m/s2

~Sub into Eq to find Force Tension on da String Son !~

5(1.16) + 5(9.8)Sin60 = FT

FT = 48.2 N
The results looks correct.

Comment: If I were grading this, I would be concerned that you are using the same variable names for mass 1 quantities and mass 2 quantities, particularly Fg and m .
 
SammyS said:
The results looks correct.

Comment: If I were grading this, I would be concerned that you are using the same variable names for mass 1 quantities and mass 2 quantities, particularly Fg and m .

Thanks for the input, I will make amends in future questions.
 
Inferior Mind said:
Thanks for the input, I will make amends in future questions.
To be more specific: FT and a are the same in magnitude for both masses, so it's fine to use the same variable name for them.
 
Last edited:
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Back
Top