Question

The leg and cast in the figure below weigh 280 N (w₁). Determine the weight w₂ and the angle 𝛼 needed so that no force is exerted on the hip joint by the leg plus cast

w₂ = ___ N

𝛼=  ___°

Answer 1

Question:

The leg and cast in the figure below weigh 280N (w₁). Determine the weight w₂ and the angle 𝛼 needed so that no force is exerted on the hip joint by the leg plus cast.

Solution:

Given:

$T_3 = w_1 = 220N\\ T_1 = 110 N\\ \theta_1 = 40°$

$\theta_2 =?\\ T_2 = w_2=?$

The representation of the image is;

Newton’s 3rd Law of Motion has already been used to isolate the joint from its physical connections by way of action-reaction pair.

Newton’s 2nd Law of Motion Use components. The x axis is the horizontal and the y axis the vertical directions.

Because there is no horizontal displacement

$\Sigma F_x = 0\\ –T_{1x} + T_{2x} = -T_1cos\theta_1 + T_2cos\theta_2 = 0$

Because Upward force is equal to downward force

$\Sigma Fy = 0 T_{1y} + T_{2y} – W_1 = 0 \\T_1sin\theta_1 + T_2sin\theta_2 – W_1 = 0$

T₂ and $\theta_2$ are both unknown. They can be eliminated. These two variables occur together as $T_2sin\theta_2$ and $T_2cos\theta_2$. This suggests solving for the angle $\theta_2$ by solving for;

$\tan\theta_2 = \dfrac{T_2\sin\theta_2}{T_2\cos\theta_2}= \dfrac{W_1 -T_1\sin\theta_1}{T_1\cos\theta_1}$

Then the angle follows from;

$\theta_2 =\alpha=\tan^{-1} \dfrac{W_1 -T_1\sin\theta_1}{T_1\cos\theta_1}$

$=\tan^{-1} \dfrac{220 -110\sin40°}{T_1\cos40°}$

$=\tan^{-1} \dfrac{149.3}{84.3}= 60.6°$

$T_2 =w_2 = T_1\dfrac{\cos\theta_1}{\cos\theta_2}= 110\dfrac{\cos40°}{\cos60.6°}$

$= 172N$