Respuesta :
According to Newtons third law vertical components of T₁ and T₂ should be equal to the weight of the box or:
T₁sin(θ₁) + T₂sin(θ₂)=200 N
Since θ₂=65°, T₁=132.6 N and T₂=130 N we need to solve for θ₁:
T₂sin(θ₂)=200 N - T₁sin(θ₁)
sin(θ₂)=(200-T₁sin(θ₁))/T₂
θ₂=sin⁻¹{(200-T₁sin(θ₁))/T₂}=37.9°
So the angle θ₂=37.9°.
T₁sin(θ₁) + T₂sin(θ₂)=200 N
Since θ₂=65°, T₁=132.6 N and T₂=130 N we need to solve for θ₁:
T₂sin(θ₂)=200 N - T₁sin(θ₁)
sin(θ₂)=(200-T₁sin(θ₁))/T₂
θ₂=sin⁻¹{(200-T₁sin(θ₁))/T₂}=37.9°
So the angle θ₂=37.9°.
Answer:
38.29
Explanation:
First you need to find the vertical component of T2 by using trig, sin65 = y/130 ... y = 130*sin65 then y for T2 is 117.8, so 200 - 117.8 tells us the vertical component of the tension of T1 which is 82.18. We can then use trig on the T1 triangle to solve for theta 1 which I will call X here....sinX= 82.17/132.6 and therefore X= sin-1 (82.17/132.6) and X (theta 1) = 38.29