Answer:
The tension in the final thread is 20.41 N and the angle is 51.73°
Explanation:
Given that,
First pulley at 90°
Second pulley at 300°
Tension on first pulley = 15 N
Tension on second pulley = 10 N
We need to calculate the tension in the final thread
Using formula of resultant
[tex]T=\sqrt{(T_{1}-T_{2}\sin\theta)^2+(T_{2}\cos\theta)^2}[/tex]
[tex]T=\sqrt{(15-10\sin60)^2+(10\cos 60)^2}[/tex]
[tex]T=20.41\ N[/tex]
We need to calculate the angle
Using formula of angle
[tex]\tan\theta=\dfrac{T_{1}-T_{2}\sin\theta}{T_{2}\cos\theta}[/tex]
Put the value into the formula
[tex]\theta=\tan^{-1}\dfrac{15-10\sin60}{10\cos 60}[/tex]
[tex]\theta=51.73^{\circ}[/tex]
The pulley should be placed at 51.73°.
Hence, The tension in the final thread is 20.41 N and the angle is 51.73°
