Answer:
[tex]T_{1} = \frac{mg cos \theta1}{sin (\theta1+\theta2)}[/tex]
Step-by-step explanation:
The angle made by strings 1 and 2 are [tex]\theta1[/tex] and [tex]\theta2[/tex] 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 [tex]T1sin\theta1[/tex] which acts upwards
Horizontal component of T1 is [tex]T1cos\theta1[/tex] which acts to the left
Vertical component of T2 is [tex]T2sin\theta2[/tex] which acts upwards
Horizontal component of T2 is [tex]T2cos\theta2[/tex] 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
[tex]T1sin\theta1+ T2sin\theta2= mg[/tex]
For horizontal forces (left and right side, there's no motion)
[tex]T1cos\theta1= T2cos\theta2[/tex]
Re-arranging the above equation and making T2 the subject of the formula we deduce that;
[tex]T2= \frac{T1cos \theta1}{cos \theta2} [/tex]
Substituting the above T2 into the vertical component we get
[tex]T1sin \theta1+ \frac{T1cos \theta1}{cos \theta2}sin \theta2=mg[/tex]
Multipyling both sides by [tex]cos\theta2[/tex] we obtain
[tex]T1sin\theta1cos\theta2+T1cos\theta1 sin\theta2 = mgcos \theta2[/tex]
[tex]T1(sin\theta1cos\theta2+cos\theta1 sin\theta2)= mgcos \theta2[/tex]
Using the concept that sin(a+b)=sin(a)cos(b)+sin(b)cos(a) we get
[tex]T1 sin (\theta1+\theta2)= mgcos\theta2[/tex]
Making T1 the subject of the formular in the above, we obtain
[tex]T1=\frac{mgcos \theta2}{sin (\theta1+\theta2)} [/tex]
Keywords: Cable, acceleration due to gravity
The expression for the tension [tex]T_{1}[/tex] is [tex]T_{1} = \left(\frac{\cos \theta_{2}}{\cos\theta_{1}} \right)\cdot \frac{m\cdot g}{\tan \theta_{1}\cdot \cos \theta_{2}+\sin \theta_{2}}[/tex]
The Newton's laws of motion models the forces acting on the system. In this question we must determine the tensions experimented by the object on equilibrium.
By Newton's laws of motion we have the following system of linear equations:
[tex]\sum F_{x} = -T_{1}\cdot \cos \theta_{1} + T_{2}\cdot \cos \theta_{2} = 0[/tex] (1)
[tex]\sum F_{y} = T_{1}\cdot \sin \theta_{1} + T_{2}\cdot \sin \theta_{2} - m\cdot g = 0[/tex] (2)
By (1):
[tex]T_{1} = T_{2} \cdot \left(\frac{\cos \theta_{2}}{\cos \theta_{1}} \right)[/tex]
(1) in (2):
[tex]T_{2} = \frac{m\cdot g}{\tan\theta _{1}\cdot \cos \theta_{2}+\sin \theta_{2}}[/tex]
And by (1):
[tex]T_{1} = \left(\frac{\cos \theta_{2}}{\cos\theta_{1}} \right)\cdot \frac{m\cdot g}{\tan \theta_{1}\cdot \cos \theta_{2}+\sin \theta_{2}}[/tex]
The expression for the tension [tex]T_{1}[/tex] is [tex]T_{1} = \left(\frac{\cos \theta_{2}}{\cos\theta_{1}} \right)\cdot \frac{m\cdot g}{\tan \theta_{1}\cdot \cos \theta_{2}+\sin \theta_{2}}[/tex]. [tex]\blacksquare[/tex]
The statement is incomplete and poorly formatted. Complete and correct statement is described below:
A chandelier with mass [tex]m[/tex] is attached to the ceiling of a large concert hall by two cables. Because the ceiling is covered with intricate architectual decorations (not indicated in the figure, which uses a humbler depiction), the workers who hung the chandelier could not attach the cables to the ceiling directly above the chandelier. Instead, they attached the cables to the ceiling near the walls. Cable 1 has tension [tex]T_{1}[/tex] and makes an angle [tex]\theta_{1}[/tex] with the ceiling. Cable 2 has tension [tex]T_{2}[/tex] and makes an angle [tex]\theta_{2}[/tex] with the ceiling.
Find an expression for [tex]T_{1}[/tex], the tension in cable 1, that does not depend on [tex]T_{2}[/tex]. Express your answer in terms of some or all of the variables [tex]m[/tex], [tex]\theta_{1}[/tex] and [tex]\theta_{2}[/tex], as well as the magnitude of the acceleration due to gravity [tex]g[/tex]. You must use parenthesis around [tex]\theta_{1}[/tex] and [tex]\theta_{2}[/tex], when they are used as arguments to any trigonometric functions in your answer.
To learn more on Newton's laws of motion, we kindly invite to check this verified question: https://brainly.com/question/13678295
