Question

A heavy mirror that has a width of 2 m is to be hung on a wall as shown in the figure below. The mirror weighs 700 N and the wire used to hang it will break if the tension exceeds 700 N (breaking strength). What is the shortest length of wire L that can be used to hang the mirror without the wire breaking? (Ignore any friction forces between the mirror and the wall.)

Answer 1

The translational equilibrium condition allows finding that the response for cable length with a maximum tension is

      L = 2.56 m

Newton's second law says that the force is proportional to the mass and the acceleration of the body, in the special case that the acceleration is zero, the relationship is called the translational equilibrium condition.

              ∑ F = 0

 

Where the bold indicates vectors, F is the force and the sum is for all external forces.

The reference systems are coordinate systems with respect to which the decomposition of the vectors is carried out and the measurements are made, in this case we will use a system with the horizontal x axis and the vertical y axis.

In the attachment we can see a free body diagram of the system, let's write the equilibrium condition for each axis.

x-axis  

         Tₓ -Tₓ = 0

y-axis

         $T_y + T_y - W =0$

         2$T_y$ - W = 0

Let's use trigonometry to decompose the tension, we can see from the graph and the adjoint that each string is half the length, let's call the angle θ

          cos θ = $\frac{T_x}{T}$

          sin θ = $\frac{T_y} { T}$

          Tₓ = T cos θ

          $T_y$ = T sin θ

We substitute

          2 T sin θ = W          (1)

The text indicates that the length of the block is 2 m, so the distance to the midpoint is

        x = 1 m

Let's use the Pythagoras' Theorem            

            H² = CA² + CO²

           CO = $\sqrt{H^2 - CA^2}$

           CA = x

           CO = $\sqrt{(\frac{L}{2} )^2 - 1 }$

Where CO is the opposite leg,  CA is the adjacent leg and H is the hypotenuse indicating H = L / 2,

Let's write the trigonometry functions

           sin θ = $\frac{CO}{H}$

Let's substitute        

            sin θ = $\frac{2 \sqrt{\frac{L^2}{4} -1 } }{L}$

Let's subtitute in the equation  1

          2 T  ( \frac{2 \sqrt{\frac{L^2}{4} -1 } }{L}   ) = W

          $\sqrt{\frac{L^2}{4}-1 } = \frac{1}{4} \ \frac{W}{T} \ L$

Let's solve by squaring

         $\frac{L^2}{4} -1 = \frac{1}{4} \ (\frac{W}{T} )^2 \ \frac{L^2}{4}$  

         $\frac{L^2}{4} ( 1 - \frac{1}{4} (\frac{W}{T})^2 ) -1 =0$

They indicate that the maximum tension of the cable is T = 700N and the weight is worth W = 700N, we substitute the values

        $\frac{L^2}{4} ( 1- \frac{1}{4}) - 1 =0 \\\frac{3 L^2}{16} = 1 \\ L^2 = \frac{16}{3} \\L = \sqrt{\frac{16}{3} }$

        L =   2.31 m    

In conclusion using the translational equilibrium condition we can find that the response for cable length with maximum tension is

      L = 2.31 m

Learn more about translational equilibrium here:

brainly.com/question/12966823

Answer 2

Answer:

2tsin(theta)-700=0

2(700)sin(theta)=700

sin (theta=(700/1400)

theta =30 degrees

cos (theta)=1/2L

L=2/cos (theta)

L=2.31m

Explanation: