How Does a Pulley System Affect Tension on an Inclined Plane?

[pikapika1]

A box of mass m2 = 1 kg on a frictionless plane inclined at angle θ = 30°. It is connected by a cord of negligible mass to a box of mass m1 = 3 kg on a horizontal frictionless surface. The pulley is frictionless and massless.

m1 O
---------------------\
\
-------------------30 \ m2
----------------------- \

(a) If the magnitude of horizontal force is 2.3 N, what is the tension in the connecting cord?

(b) What is the largest value the magnitude of may have without the cord becoming slack?

m1
F+T=m1a
2.3+T =3a

m2
sin30 m2g - T = m2a
4.9-T=a

substitution
2.3+T = 3(4.9-T)
4T = 12.4-2.3
T = 3.1N

I'm not sure if this is correct, can somebody verify this for me and give me a hint on how to do part b)?

thanks in advance

 

[alphysicist]

Hi pikapika1,

A box of mass m2 = 2.60 kg on a frictionless plane inclined at angle θ = 33°. It is connected by a cord of negligible mass to a box of mass m1 = 4.00 kg on a horizontal frictionless surface. The pulley is frictionless and massless.

m1 O
---------------------\
\
30\ m2
----------------------- \

(a) If the magnitude of horizontal force is 2.3 N, what is the tension in the connecting cord?

(b) What is the largest value the magnitude of may have without the cord becoming slack?

m1
F+T=m1a

I don't think it was clear from your post about what the horizontal force was. Based on this equation, I'm guessing that the horizontal force is acting on m1 and is directed toward the incline; is that correct?

2.3+T =3a

m2
sin30 m2g - T = m2a
4.9-T=a

This line seems to be wrong; it is missing the m₂ from the previous line.

 

[pikapika1]

the horizontal force is applied on m1

and since m2 = 1kg
i simplified it to m2a = a

 

[alphysicist]

the horizontal force is applied on m1

and since m2 = 1kg
i simplified it to m2a = a

What happened to the original problem? As you can see from what I quoted in my previous post, you had m2=2.6 kg, a different m1, and a different angle.

 

[pikapika1]

What happened to the original problem? As you can see from what I quoted in my previous post, you had m2=2.6 kg, a different m1, and a different angle.

yea sorry i realized that my numbers were messed up. Could you help me with the latter numbers that i have given?

 

[alphysicist]

yea sorry i realized that my numbers were messed up. Could you help me with the latter numbers that i have given?

The tension you found (3.1 N) looks correct to me for those numbers.

For part b, if the rope has just barely become slack, what does that indicate about the tension in the rope and the accelerations of the two objects?

 

[pikapika1]

so find F when T = 0?

 

[alphysicist]

so find F when T = 0?

That's right; and also the objects are still moving at the same rate. What do you get?

 

[pikapika1]

so find F when T = 0?

f= m1a
a= f/m1

m2gsin30=m2a
m2gsin30*m1/(m2) = f
f= 14.7N

 

[alphysicist]

so find F when T = 0?

f= m1a
a= f/m1

m2gsin30=m2a
m2gsin30*m1/(m2) = f
f= 14.7N

That looks right to me.