Joe wishes to hang a sign weighing 750 N so that cable A, attached to the store, makes a 30 degree angle. Cable B is horizontal and attached to an adjoining building. What is the tension in cable B, work please.
2006-10-31 12:07:56 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Cable B is not providing a vertical component, so cable A has to provide all the vertical force to support the sign. The vertical component of its force on the sign is 750N. This value is also equal to the tension in cable A times sin 30 deg. Knowing that sin 30 deg is 0.5, you can solve for the tension in cable A.
The horizontal component of the force that cable A exerts on the sign has to be offset by the horizontal force of cable B, which is equal to the tension in cable B. You can calculate the horizontal force from cable A as the tension in cable A times cos 30 deg. Knowing that cos 30 deg is sqrt(3)/2, you can calculate the tension in cable B.
2006-10-31 12:25:16 · answer #1 · answered by actuator 5 · 0⤊ 0⤋
750/sin150=T/sin120
t=750*1/2*sqrt(3)/2=750*sqrt(3)/4
2006-11-01 14:28:57 · answer #2 · answered by Anonymous · 0⤊ 0⤋