Question

Find an expression for T1, the tension in cable 1, that does not depend on T2. Express your answer in terms of some or all of the variables m, θ1, and θ2, as well as the magnitude of the acceleration due to gravity g. You must use parentheses around θ1 and θ2, when they are used as arguments to any trigonometric functions in your answer.

Answer 1

Answer:

$T_{1} = \frac{mg cos \theta1}{sin (\theta1+\theta2)}$

Step-by-step explanation:

The angle made by strings 1 and 2 are $\theta1$ and $\theta2$ respectively. Tension made by strings 1 and 2 are T1 and T2 respectively. Considering the diagram shown, the following tension components are available;

Weight of the block mg acting on the body directing the force downwards

Vertical component of T1 is $T1sin\theta1$ which acts upwards

Horizontal component of T1 is $T1cos\theta1$ which acts to the left

Vertical component of T2 is   $T2sin\theta2$ which acts upwards

Horizontal component of T2 is  $T2cos\theta2$ acting towards right

To maintain a state of equilibrium, the upward forces and downwards forces must be equal even as the forces on the right and left hand side must balance

Therefore, balancing the vertical components we obtain

$T1sin\theta1+ T2sin\theta2= mg$

For horizontal forces (left and right side, there's no motion)

$T1cos\theta1= T2cos\theta2$

Re-arranging the above equation and making T2 the subject of the formula we deduce that;

$T2= \frac{T1cos \theta1}{cos \theta2}$

Substituting the above T2 into the vertical component we get

$T1sin \theta1+ \frac{T1cos \theta1}{cos \theta2}sin \theta2=mg$

Multipyling both sides by $cos\theta2$ we obtain

$T1sin\theta1cos\theta2+T1cos\theta1 sin\theta2 = mgcos \theta2$

$T1(sin\theta1cos\theta2+cos\theta1 sin\theta2)= mgcos \theta2$

Using the concept that sin(a+b)=sin(a)cos(b)+sin(b)cos(a) we get

$T1 sin (\theta1+\theta2)= mgcos\theta2$

Making T1 the subject of the formular in the above, we obtain

$T1=\frac{mgcos \theta2}{sin (\theta1+\theta2)}$

Keywords: Cable, acceleration due to gravity