A 215 N sign is supported by two ropes. One rope pulls up and to the right 1=29.5∘ above the horizontal with a tension 1 , and the other rope pulls up and to the left 2=44.5∘ above the horizontal with a tension 2 , as shown in the figure. Find the tensions 1 and 2 .

The sign is held in equilibrium. Using Newton's second law, we set up the equations of the net forces acting on the sign in the horizontal and vertical directions:

∑ F (horizontal) = T₁ cos(29.5°) - T₂ cos(44.5°) = 0

(right is positive, left is negative)

∑ F (vertical) = T₁ sin(29.5°) + T₂ sin(44.5°) - 215 N = 0

(up is positive, down is negative)

Solve the system of equations. I use elimination here:

• Multiply the first equation by sin(29.5°) and the second by cos(29.5°):

sin(29.5°) (T₁ cos(29.5°) - T₂ cos(44.5°)) = 0

cos(29.5°) (T₁ sin(29.5°) + T₂ sin(44.5°) - 215 N) = 0

T₁ cos(29.5°) sin(29.5°) - T₂ cos(44.5°) sin(29.5°) = 0

T₁ cos(29.5°) sin(29.5°) + T₂ cos(29.5°) sin(44.5°) = (215 N) cos(29.5°)

• Subtract the first equation from the second to eliminate T₁ :

T₂ cos(29.5°) sin(44.5°) - (- T₂ cos(44.5°) sin(29.5°)) = (215 N) cos(29.5°)

• Solve for T₂ :

T₂ (cos(29.5°) sin(44.5°) + cos(44.5°) sin(29.5°)) = (215 N) cos(29.5°)

T₂ sin(74.0°) = (215 N) cos(29.5°)

… … … (using the fact that sin(x + y) = sin(x) cos(y) + cos(y) sin(x))

T₂ = (215 N) cos(29.5°) / sin(74.0°)

T₂ ≈ 195 N

• Solve for T₁ :

T₁ cos(29.5°) - T₂ cos(44.5°) = 0

T₁ cos(29.5°) = T₂ cos(44.5°)

T₁ = T₂ cos(44.5°) / cos(29.5°)

T₁ ≈ 160. N

explanation:

newton's third law, for every action there is an equal and opposite reaction.

this could be newton#'s law of cooling. the cooling curve is exponential