Question

A traffic light of weight w hangs from two lightweight cables, one on each side of the light. Each cable hangs at a 45° angle from the horizontal. What is the tension in each cable ?
w/2 (ii) w/√2 (iii) w√2 (iv) w (v) 2w
Find an equation of the tangent line to the parabola at the
point .

Answer 1

Both the strings will have T.sin45° value of vertical component of tension

Answer 2

Answer:

The tension in each cable is (iii) $w/\sqrt{2}$

Explanation:

The free-body diagram for this situation is given below.

Step 1:

Given the weight of the traffic light to be w, the angle subtended by the cables with the horizontal is $45^{\circ$. From the diagram $\angle FCB=\angle CFD = 45^\circ$.

The gravitational force acts vertically down.

Step 2:

Using the property that vertically opposite angles are equal we can write

                                     $\angle ABC=\angle EDF= 45^\circ$

Resolving the tension in the cable in the horizontal and vertical directions we get,

$Tsin\theta$ points vertically up for the 2 cables

$Tcos\theta$ for the left cable is horizontally towards the left and for the right cable horizontally towards the right.

Step 4:

Here there is no motion in either the vertical or the horizontal directions. Hence from Newton's 2nd Law, net force along the 2 directions are zero.

In the horizontal direction, the forces are equal and opposite hence cancelling each other.

In the vertical direction, the force acting upwards due to the two cables is $2Tsin\theta$ and is balanced by the weight of the traffic light. Hence,

                                    $2Tsin\theta=w$

Substituting for $\theta$ we have,

                                    $2Tsin45^\circ=w\\\frac{2T}{\sqrt{2} } =w\\T=\frac{w}{\sqrt{2} }$

Therefore from the above equation, we get the tension due to each cable being $\frac{w}{\sqrt{2} }$ given in option (iii).