A 331 N stoplight is hanging in equilibrium from cables as shown. The tension in the right cable is 550 N, and it makes an angle
of 12 degrees with the horizontal. What is the force of tension exerted by the cable on the left side? 114 N 217 N 538 N 580 N
1 answer:
Let
F=tension of cable on the left
x=angle of left cable makes with the horizontal
Resolve forces vertically (y-direction)
331=550sin(12)+Fsin(x) =>
Fsin(x)=331-550sin(12)=216.649..........(1)
Resolve forces horizontally (x-direction)
Fcos(x)=550cos(12) =>
Fcos(x)=537.981............................(2)
Divide (1) by (2)
tan(x)=Fsin(x)/Fcos(x)=0.40271
x=atan(0.40271) = 21.935 °
From (2),
F=537.981/cos(x)=537.981/0.92761 = 579.965 N
Check:
Fsin(x)+550sin(12)=579.965*sin(21.935)+550*sin(12) = 331.000 ok
Fcos(x)=537.981
550sin(12)=537.981 ok
You might be interested in
Fusion and gravity is ur answer
Answer:
Explanation:
Let r be the rate of the slower walker in mph
[r + (r + 1.7)](2) = 13
(2r + 1.7)(2) = 13
4r + 3.4 = 13
4r = 9.6
r = 2.4 mph
r + 1.7 = 4.1 mph
Cathode ray tube, plumb pudding model, atoms were not hard spheres, but had smaller parts.
The atom's atomic number is the number of protons in its nucleus.
Answer:
Explanation:
Proton is the correct answer