Question

A 153.8-N bird feeder is supported by three cables as shown in the figure below. Find the tension in each cable.left cable _____ Nright cable _____ Nbottom cable _____ N

Answer 1

Given:

The weight of the bird feeder is: W = 153.8 N

To find:

The tension in each cable.

Step-by-step explanation:

Let T represent the tension in the cable attached to the bird feeder i.e, in the bottom cable, T1 be the tension in the left cable and T2 be the tension in the right string.

As the system is at equilibrium, the tension T in the cable attached to the bird feeder will be equal to the weight of the bird feeder. Thus,

T = W = 153.8 N

The components of tension T1 in the left cable and tension T2 in the right cable are resolved as shown in the diagram.

From the diagram, by considering horizontal components, we get:

`\begin{gathered} T_1cos60\degree=T_2cos30\degree \\ \\ T_1=(T_2cos30\degree)/(cos60\degree)..........(1) \end{gathered}`

From the diagram, by considering the vertical components, we get:

`\begin{gathered} T_1sin60\degree+T_2sin30\degree=W \\ \\ T_1s\imaginaryI n60\degree+T_2s\imaginaryI n30\degree-W=0 \\ \\ T_1s\imaginaryI n60\degree+T_2s\imaginaryI n30\degree-153.8\text{ N}=0..........(2) \end{gathered}`

Substituting equation (1) in equation (2), we get:

`\begin{gathered} (T_2cos30\degree)/(cos60\degree)* sin60\degree+T_2sin30\degree-153.8\text{ N}=0 \\ \\ T_2cos30\degree tan60\degree+T_2sin30\degree-153.8\text{ N}=0 \\ \\ T_2*(√(3))/(2)*√(3)+T_2*(1)/(2)=153.8\text{ N} \\ \\ (3)/(2)T_2+(1)/(2)T_2=153.8\text{ N} \\ \\ 2T_2=153.8\text{ N} \\ \\ T_2=\frac{153.8\text{ N}}{2} \\ \\ T_2=76.9\text{ N..........\lparen3\rparen} \end{gathered}`

Substituting equation (3) in equation (1), we get:

`\begin{gathered} T_1=\frac{76.9\text{ N}*((√(3))/(2))}{((1)/(2))} \\ \\ T_1=76.9\text{ N}*√(3) \\ \\ T_1=133.2\text{ N} \end{gathered}`

Final answer:

The tension in the left cable is 133.2 Newtons.

The tension in the right cable is 76.9 Newtons.

The tension in the bottom cable is 153.8 Newtons.