Question

Find the value of T₁ and T2 in the figure below. A t 27° T₁ B BRAJE 38° T₂ C 12g N Hint: Apply triangle of forces rule or sine and cosine rules

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Answer 1

T₁ = 102.25 N (2 d.p.)

T₂ = 115.61 N (2 d.p.)

Step-by-step explanation:

The diagram shows a body of mass 12 kg (weight = 12g N) held in equilibrium by two light, inextensible strings. One string makes an angle of 27° with the positive horizontal and the other string makes an angle of 38° with the negative horizontal.

The body is in equilibrium, so both the horizontal and vertical components of the forces must sum to zero.

Resolving horizontally, taking (→) as positive:

`\implies -T_1 \cos (27^(\circ))+T_2 \cos (38^(\circ))=0`

`\implies T_1 \cos (27^(\circ))=T_2 \cos (38^(\circ))`

`\implies T_1=(T_2 \cos (38^(\circ)))/( \cos (27^(\circ)))`

Resolving vertically, taking (↑) as positive:

`\implies T_1 \sin(27^(\circ))+T_2 \sin(38^(\circ))-12\text{g}=0`

`\implies T_1 \sin(27^(\circ))+T_2 \sin(38^(\circ))=12\text{g}`

Substitute the found expression for T₁ into the second equation and take g = 9.8 ms⁻²:

`\implies \left((T_2 \cos (38^(\circ)))/( \cos (27^(\circ)))\right) \sin(27^(\circ))+T_2 \sin(38^(\circ))=12\text{g}`

`\implies T_2 \cos (38^(\circ))\tan(27^(\circ))+T_2 \sin(38^(\circ))=12\text{g}`

`\implies T_2 \left(\cos (38^(\circ))\tan(27^(\circ))+ \sin(38^(\circ))\right)=12\text{g}`

`\implies T_2 =\frac{12\text{g}}{\cos (38^(\circ))\tan(27^(\circ))+ \sin(38^(\circ))}`

`\implies T_2=115.614550...\:\text{N}`

Substitute the found value of T₂ into the equation for T₁ and take g = 9.8 ms⁻²:

`\implies T_1=\frac{\left(\frac{12\text{g}}{\cos (38^(\circ))\tan(27^(\circ))+ \sin(38^(\circ))}\right) \cos (38^(\circ))}{ \cos (27^(\circ))}`

`\implies T_1=102.250103...\text{N}`

Therefore, the value of T₁ and T₂ in the given diagram is:

• T₁ = 102.25 N (2 d.p.)
• T₂ = 115.61 N (2 d.p.)

Answer 2

The tension in
`T_1` is 0.1024 Newtons.

The tension in
`T_2` is 0.1157 Newtons.

Step-by-step explanation:

Lets create a free body diagram showing all of the forces; we need to show the vertical and horizontal components of the tension.

I will attach a picture of my free body diagram. Notice I created 2 new triangles with the adjacent angles of angle A and C from the original picture.

Lets make a list of all the variables we have now. Also lets write down the information we are given.

`A=27\\C=38\\m=12\\g=9.81\\W\\T_(1) \\T_(1x) \\T_(1y)\\T_(2) \\T_(2x) \\T_(2y) \\`

In this situation the sum of of the vertical tension components must support the weight. To find the vertical components we can use the SIN function.

`sin(x)=(O)/(H) \\O=Hsin(x)`

Therefore we can write that the sum of the forces in the y direction is

`\sum F_y=T_1sin(A)+T_2sin(C)=W`

This system is in equilibrium; the object should not move along the x-axis. Therefore, the horizontal components of
`T_1` and
`T_2` must then equal each other. To find the horizontal components we can use the COS function.

`cos(x)=(A)/(H) \\A=Hcos(x)`

Therefore we can write that the sum of the forces in the x direction is

`\sum F_x=T_1cos(A)=T_2cos(C)`

Now we have to equations to help us solve the problem.

`T_1cos(A)=T_2cos(C)`

`T_1sin(A)+T_2sin(C)=W`

We do not know the numerical values of
`T_1` and
`T_2` so we will have to manipulate algebraically to solve them.

In the first equation lets solve for
`T_1`.

`T_1cos(A)=T_2cos(C)`

Divide both sides by
`cos(A)`.

`T_1=(T_2cos(C))/(cos(A))`

Separate the right side into two fractions.

`T_1=(T_2)/(1) *(cos(C))/(cos(A))`

Use the reciprocal trig identity for cosine.

`sec(x)=(1)/(cos(x))`

`T_1=(T_2)/(1) *cos(C)}*sec(A)`

`T_1=T_2*cos(C)}*sec(A)`

Now insert our answer for
`T_1` into the second equation.

`T_2*cos(C)}*sec(A)*sin(A)+T_2*sin(C)=W`

Solve for
`T_2`. Lets replace each trig function with its own variable to make this easier.

`T_2*cos(C)}*sec(A)*sin(A)+T_2*sin(C)=W`

`cos(C)=x\\sec(A)=y\\sin(A)=z\\sin(C)=u`

`T_2*x*y*z+T_2*u=W\\T_2xyz+T_2u=W`

Now lets solve for
`T_2`.

Factor
`T_2` out of each term.

`T_2(xyz)+T_2(u)=W`

Factor
`T_2` out of each term.

`T_2(xyz+u)=W`

Divide each side by
`xyz+u`.

`T_2=(W)/(xyz+u)`

Lets substitute the trig functions back in for the variables

`T_2=(W)/(cos(C)*sec(A)*sin(A)+sin(C))`

`W` is the weight.

The formula for weight is
`W=mg`. Where
`m` is the mass in kilograms.

12 grams is 0.012 kilograms.

`W=0.012*9.81`

`W=0.11772`

Numerical Evaluation

Lets evaluate
`T_2`.

`T_2=(0.11772)/(cos(38)*sec(27)*sin(27)+sin(38))`

`T_2=0.11573`

Lets evaluate
`T_1`.

`T_1cos(A)=T_2cos(C)`

`T_1*cos(27)=0.11573*cos(38)\\T_1=0.10235443`

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