91影视

Given that a holiday decoration consists of two shiny glass spheres with masses 0.0240 kg and 0.0360 kg suspended from a uniform rod with mass 0.120 kg and length 1.00 m. The rod is suspended from the ceiling by a vertical cord at each end, so that it is horizontal. Calculate the tension in each of the cords A through F.

Mass of sphere hanging from A, $m_1=0.036\mathrm{kg}$

Mass of sphere hanging from B, $m_2=0.024\mathrm{kg}$

Let the tension in the cords A, B, C, D, E, F respectively are$T_A,T_B,T_C,T_D,T_E,T_F$ .

Consider the formula for the torque is

$\mathbf{蟿}\mathbf{=}\mathbf{Fl}$

Here, Fis force exerted and l is moment arm.

Tension in cord A is simply the weight hanged.

Mass hanged on A is $m_1$

So Tension ${\mathrm{T}}_{\mathrm{A}}=\left(0.036\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)=0.353\mathrm{N}$

Hence, tension in chord A is 0.353 N.

Tension in cord B is simply the weight hanged.

Mass hanged on B is$m_1+m_2$

So Tension

Hence, tension in chord A is 0.588 N.

Tension in cord C can be found using condition of equilibrium

Apply$\underset{}{鈭�}{F}_x=0$ and$\underset{}{鈭�}{F}_y=0$ to the point where cords are joined.

Tension in cord C $T_C=T_B\cos 36.9掳$

So Tension ${\mathrm{T}}_{\mathrm{C}}=\left(0.588\mathrm{N}\right)\cos 36.9掳=0.47\mathrm{N}$

Hence, tension in chord C is 0.47 N.

Tension in cord D can be found using condition of equilibrium

Apply $\underset{}{鈭�}{F}_x=0$and $\underset{}{鈭�}{F}_y=0$to the point where cords are joined.

Tension in cord D$T_D=T_B\cos 53.1掳$

So Tension $T_D=\left(0.588\mathrm{N}\right)\cos 53.1掳=0.353N$

Hence, tension in chord D is 0.353 N.

Tension in cord E can be found by taking torques about the point where string F is attached.

So $T_E\left(1\mathrm{m}\right)=T_D\sin 鈦�{36.9}^{鈭�}\left(0.8\mathrm{m}\right)+T_C\sin 鈦�{53.1}^{鈭�}\left(0.2\mathrm{m}\right)+\left(0.12\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(0.5\mathrm{m}\right)$

So Tension $T_E=0.833N$

Hence, tension in chord E is 0.833 N.

Now,$T_E+T_F$ must be total weight of the ornament.

So $T_E+T_F=\left(0.18kg\right)\left(9.8m/s^2\right)$

So Tension $T_F=0.931N$

Hence, tension in chord A is 0.931 N.