I am trying to express some force F_h in terms of mg but the reuslt I get is wrong the equations are the following:
$$\Sigma F_x = N_2-f1 = 0$$ $$\Sigma F_y = f2 + N_1-mg-F_h = 0$$
$$\Sigma\tau = -\frac{1}{2}Lmg\sin\theta - LF_h\sin\theta + LN_2\cos\theta+Lf_2\sin\theta = 0$$
notice how using the first two equation I express N_2 in terms of mg and F_h hence:
$$N_2 = \frac{\mu(mg + F_h)}{\mu^2+1}$$
solving for F_h in the third equation and plug in the new N_2 eq I get:
$$F_h =mg \frac{-\frac{1}{2}-\frac{\mu \cot\theta+\mu^2}{\mu^2+1}}{1 +\frac{\cot\theta-\mu}{\mu^2+1}}$$
since the angle is 40° and μ is 0.4 I get:
$$F_h = mg0.45$$
however this is wrong the correct answer is: $$F_h = mg0.11$$
what did I get wrong?
for the context this equations are the static equilibrium of a meter stick with some upper angle 40° and μ 0.4 and I am trying to see how hard can I press down vertically on the top of the stick with a hand before slipping begins thats why I am trying to solve for F_h force of hand. notice how I use the point of contact of the stick with the ground as pivot in the torque equation.
Your expression for $N_2$ is correct, but there are some algebra mistakes when substituting it in and solving for $F_h$. Notice that: $$ F_h = \frac{LN_2 \cos \theta + L(\mu N_2)\sin\theta - 0.5Lmg\sin\theta}{L\sin\theta} = (\cot\theta + \mu)N_2 - 0.5mg $$ Substituting, we have: \begin{align*} F_h &= (\cot\theta + \mu) \cdot \frac{\mu}{\mu^2 + 1}(mg + F_h) - 0.5mg \\ F_h &= \frac{\mu(\cot\theta + \mu)}{\mu^2 + 1}(mg + F_h) - 0.5mg \\ F_h - \frac{\mu(\cot\theta + \mu)}{\mu^2 + 1}F_h &= \frac{\mu(\cot\theta + \mu)}{\mu^2 + 1}mg - 0.5mg \\ F_h &= \frac{\dfrac{\mu(\cot\theta + \mu)}{\mu^2 + 1} - 0.5}{1 - \dfrac{\mu(\cot\theta + \mu)}{\mu^2 + 1}} mg =(0.108354\ldots)mg \end{align*}