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Forces Problem - Please help ASAP

A 32.8-kg traffic light is supported by two wires.

T1 is of the right wire which connects to the vertical pole at 30 degrees below horizontal.

T2 is of the left wire which connects to the horizontal top pole at 60 degrees below horizontal.

Is the tension in wire 2 greater than or less than the tension in wire 1?

Prove your answer by applying Newton's laws and solving for the two tensions.

Comments

Ben,

  Hmmm...I cannot picture how the poles are situated.  This needs a picture or again....1000 words!  BruceS

To solve this problem you set up two equations. One for the x component of tension and one for the y component of tension.

T1*cos 30 - T2 cos 60 = 0; X equation

T1*sin 30 + T2 sin 60 = 32.8kg * 9.8 m/s^2 Y equation

If we multiply top equation by sin 60, bottom by cos 60 and subtract we get the following:

T1cos30sin60 + T1sin30cos 60 = 32.8*9.8*cos 60

0.75*T1 + .25T1 = 160.72n = T1

substitute this value of T1 into the first equationn

.86605 * T1 = 0.5 * T2
T2 = 278.38n

 

To solve this problem you set up two equations. One for the x component of tension and one for the y component of tension.

T1*cos 30 - T2 cos 60 = 0; X equation

T1*sin 30 + T2 sin 60 = 32.8kg * 9.8 m/s^2 Y equation

If we multiply top equation by sin 60, bottom by cos 60 and subtract we get the following:

T1cos30sin60 + T1sin30cos 60 = 32.8*9.8*cos 60

0.75*T1 + .25T1 = 160.72n = T1

substitute this value of T1 into the first equationn

.86605 * T1 = 0.5 * T2
T2 = 278.38n

 

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1 Answer

To solve this problem you set up two equations. One for the x component of tension and one for the y component of tension.

T1*cos 30 - T2 cos 60 = 0; X equation

T1*sin 30 + T2 sin 60 = 32.8kg * 9.8 m/s^2 Y equation

If we multiply top equation by sin 60, bottom by cos 60 and subtract we get the following:

T1cos30sin60 + T1sin30cos 60 = 32.8*9.8*cos 60

0.75*T1 + .25T1 = 160.72n = T1

substitute this value of T1 into the first equationn

.86605 * T1 = 0.5 * T2
T2 = 278.38n