Question

A 140 kg camera is suspended by two wires over a 40 metre wide football field to get shots of the action from above. At one point, the camera is closer to the left side of the field. The tension in the wire on the left is 1500 N, while the tension in the wire on the right is 800 N. Determine the length of each cable.

Answer 1

Please red the answer below

Explanation:

In order to determine the length of each cable you use the Newton second law for each component of the forces involved in the situation.

For the x component you have:

`T_1cos\theta_1-T_2cos\theta_2=0` (1)

T1: tension of the first cable = 1500N

T2: tension of the second cable = 800N

θ1: angle between the horizontal and the first cable

θ2: angle between the horizontal and the first cable

For the y component you have:

`T_1sin\theta_1+T_2sin\theta_2-W=0` (2)

W: weight of the camera = Mg = (140kg)(9.8m/s^2) = 1372N

You can squared both equations (1) and (2) and the sum the two equations:

`T_1^2cos^2\theta_1=T_2^2cos^2\theta_2\\\\T_1^2sin^2\theta_1=T_2^2sin^2\theta_2-2WT_2sin\theta_2+W^2`

Then, you sum the equations:

`T_1^2(cos^2\theta_1+sin^2\theta_1)=T_2^2(sin^2\theta_2+cos^2\theta_2)-2Wsin\theta_2+W^2` (3)

Next, you use the following identity:

`sin^2\theta+cos^2\theta=1`

and you obtain in the equation (3):

`T_1^2=T_2^2-2WT_2sin\theta_2+W^2\\\\sin\theta_2=(T_2^2-T_1^2+W^2)/(2WT_2)=((800N)^2-(1500)^2+(1372N)^2)/(2(800N)(1372N))=0.066\\\\\theta_2=sin^(-1)0.066=27.23\°`

With this values you can calculate the value of the another angle, by using the equation (1):

`\theta_1=cos^(-1)((T_2cos(27.23\°))/(T_1))=cos^(-1)(((800N)(cos27.23\°))/(1500N))\\\\\theta_1=61.69\°`

Now, you can calculate the length of each cable by using the information about the width of the football field. You use the following trigonometric relation:

`l_1cos\theta_1=40-d\\\\l_2cos\theta_2=d\\\\`

d: distance to the right side of the field

By using the cosine law you can fins a system of equation and then you can calculate the values of l1 and l2.