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Mivz18

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Problem : Three strings, in the horizontal plane, meet in a knot and are pulled with three forces such that the knot is held stationary. The tension in string 1 is T1 = 2.5 N. The angle between strings 1 and 2 is q12 = 130° and the angle between strings 1 and 3 is q13 = 120° with string 3 below string 1 as shown.

a) Find the tension in string 2.

b) Find the tension in string 3.

A mass of 1.2 kg is now placed on the knot and supported by a frictionless table in the plane of the strings.

c) Find the acceleration of the mass if all the forces remain the same as above.

d) If the sizes and directions of T2 and T3 remain the same, but T1 is increased by 1.5 N, what is the acceleration of the mass?

Ok, first I decompose the tensions into their x and y components. After substitutions and combining equations, I end up with these final equations.

T2x = T2(cos 130)

T2y = T2(sin 130)

T3x = T3(cos (-120))

T3y = T3(sin (-120))

I can then put the formulas in a cartesian coordinate system:

(2.5 + T2(cos 130) + T3(cos -120) ) , (T2(sin 130) - T3(cos -120) ) = (0,0)

2.5 + T2(cos 130) + T3(cos -120) = 0

T2(sin 130) - T3(cos -120) = 0

T2 = (T3(sin -120)) / Sin(130)

When setting the two equations equalling zero to each other, I substituted the previous for T2. Then when solving for T3 I get 1.533 N.

Am I doing this correctly because the program I am using says that is incorrect and I am stuck.

a) Find the tension in string 2.

b) Find the tension in string 3.

A mass of 1.2 kg is now placed on the knot and supported by a frictionless table in the plane of the strings.

c) Find the acceleration of the mass if all the forces remain the same as above.

d) If the sizes and directions of T2 and T3 remain the same, but T1 is increased by 1.5 N, what is the acceleration of the mass?

Ok, first I decompose the tensions into their x and y components. After substitutions and combining equations, I end up with these final equations.

T2x = T2(cos 130)

T2y = T2(sin 130)

T3x = T3(cos (-120))

T3y = T3(sin (-120))

I can then put the formulas in a cartesian coordinate system:

(2.5 + T2(cos 130) + T3(cos -120) ) , (T2(sin 130) - T3(cos -120) ) = (0,0)

2.5 + T2(cos 130) + T3(cos -120) = 0

T2(sin 130) - T3(cos -120) = 0

T2 = (T3(sin -120)) / Sin(130)

When setting the two equations equalling zero to each other, I substituted the previous for T2. Then when solving for T3 I get 1.533 N.

Am I doing this correctly because the program I am using says that is incorrect and I am stuck.