Vectors, Tension Question: A mass of 10 kg is suspended by two pieces of string, 30 cm and 40 cm long....... from two points that are 50 cm apart and at the same level. Find the tension in each...

Vectors, Tension Question: A mass of 10 kg is suspended by two pieces of string, 30 cm and 40 cm long...

.... from two points that are 50 cm apart and at the same level. Find the tension in each piece of string.

Asked on by cm0011

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thilina-g | College Teacher | (Level 1) Educator

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Let the two points on the same level be A and B and the point from which the mass is hanging be C. Let AC =30 and BC =40.

Then according to the given lengths the triangle ABC is a right angled triangle with AB being its hypotenuse.Let the angle BAC be `alpha` with `sin(alpha) = 4/5` and `cos(alpha) = 3/5` .

If you draw the diagram and mark the angles correctly, you would see the following.

The angle AC makes with vertical = `pi/2 -alpha`

The angle AC makes with horizontal = `alpha`

The angle BC makes with vertical = `alpha`

The angle BC makes with horizontal = `pi/2 -alpha`

 

Now we can balance the horizontal and vertical components of forces at C.

The force due to the mass 10 kg is vertical and is equal to

`9.81 xx 10 = 98.1 N`

If the tension in AC is `T_(AC)` and and tension in BC is `T_(BC)`

Horizontal components,

`T_(AC) cos(alpha) = T_(BC) cos(pi/2-alpha)`

`T_(AC) cos(alpha) = T_(BC) sin(alpha)`

`T_(AC) 3/5 = T_(BC) 4/5`

Therefore,

`T_(AC) = 4/3 T_(BC)`

 

Vertical components,

`98.1 = T_(AC)cos(pi/2-alpha) +T_(BC) cos(alpha)`

`98.1 = T_(AC) sin(alpha) +T_(BC) cos(alpha)`

`98.1 = T_(AC) 4/5 +T_(BC) 3/5`

But we know that, `T_(AC) = 4/3 T_(BC)`

Therefore,

`98.1 = 4/5 xx 4/3 T_(BC)+T_(BC) 3/5`

`98.1 = 5/3 T_(BC)`

 

Therefore `T_(BC) = 58.86 N`

and `T_(AC) = 4/3 xx 58.86 N = 78.48 N`

 

The tension in AC (30 cm) is 78.48 N and tension BC (40 cm) is 58.86 N

 

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