URGENT HELP PLEASE

Question

A 500 pound object hangs from two cables with angles below the horizontal of and respectively. Determine the magnitude of the tension vectors,  and .

Answer 1

I think you are describing this:

T₂ is broken into it's two components, one in the x direction and 1 in the y direction

T_1x = -T₁cos(25°) [Note that we need to apply a negative to this because it is in the negative x-direction]

T_1y = T₁sin(25°)

T₂ is broken into it's two components, one in the x direction and 1 in the y direction

T_2x = T₂cos(32°)

T_2y = T₂sin(32°)

There is no motion so the sum of the forces in all directions is zero.

∑F_x = -T_1x + T_2x = 0

-T₁cos(25°) + T₂cos(32°) = 0

-0.9063T₁ + 0.848T₂ = 0

0.9063T₁ = 0.848T₂

T₁ = 0.848/0.9063T₂

T₁ = 0.9357T₂

∑F_y = -W + T_1y + T_2y = 0

-500 + T₁sin(25°) + T₂sin(32°) = 0

-500 + 0.4226T₁ + 0.5299T₂ = 0 (substituting in that T₁ = 0.9357T₂ from above):

-500 + 0.4226(0.9357T₂) + 0.5299T₂ = 0

-500 + 0.3955T₂ + 0.5299T₂ = 0

0.3955T₂ + 0.5299T₂ = 500

0.9254T₂ = 500

T₂ = 500/0.9254 = 540 lbs

T₁ = 0.9357T₂ = 0.9357(540) = 506 lbs

Answer 2

Break the problem down to sums of horizontal and vertical forces.

ΣF_x = T₂ cos 32^º - T₁ cos 25^º = 0.

ΣF_y = T₁ sin 25^º + T₂ sin 32^º - F_g = 0, where weight F_g = 500 pounds.

Horizontal components give us T₂ = T₁ cos 25 / cos 32 = 1.0687 T₁. Substitute into vertical equation:

0.42262 T₁ + (1.0687 T₁) (0.52992) = 500 = 0.9889 T₁.

Then T₁ = 500/0.9889 = 506 pounds, T₂ = 1.0687 T₁ = 540 pounds.

(As a check, substitute these values back into vertical equation to show that the two vertical components of tension do add up to the weight of the object)