1. **Stating the problem:** We are given forces and moments acting on a rod fixed at a clamp, with known values $M_y=4$ Nm, $M_z=2$ Nm, $F=3$ N, $F_G=1$ N, $a=1$ m, and angle $\alpha=0^\circ$. We want to find the reaction forces $A_x$, $A_y$, $A_z$ and moments $M_{Ax}$, $M_{Ay}$, $M_{Az}$ at the clamp. 2. **Equations of equilibrium:** - Sum of forces in $x$-direction: $\Sigma F_x = A_x$ - Sum of forces in $y$-direction: $\Sigma F_y = A_y + F + F_G = 0$ - Sum of forces in $z$-direction: $\Sigma F_z = A_z + F = 0$ - Sum of moments about $x$ at point $P$: $\Sigma M_x^P = M_{Ax} + mF + mF_G$ - Sum of moments about $y$ at point $P$: $\Sigma M_y^P = M_{Ay} + M_y + F + A_z$ - Sum of moments about $z$ at point $P$: $\Sigma M_z^P = M_{Az} + M_z + mF + mF_G + A_y$ 3. **Known values and simplifications:** - Since $\alpha=0^\circ$, force $F$ acts along the $y$ or $z$ axis as given. - $m$ is the moment arm length, here $a=1$ m. 4. **Calculate reaction forces:** From $\Sigma F_y=0$: $$A_y + F + F_G = 0 \implies A_y = -F - F_G = -3 - 1 = -4 \text{ N}$$ From $\Sigma F_z=0$: $$A_z + F = 0 \implies A_z = -F = -3 \text{ N}$$ From $\Sigma F_x = A_x$ and no other forces in $x$, we have: $$A_x = 0 \text{ N}$$ 5. **Calculate reaction moments:** Assuming $m=a=1$ m, From $\Sigma M_x^P=0$: $$M_{Ax} + mF + mF_G = 0 \implies M_{Ax} = -mF - mF_G = -1\times3 - 1\times1 = -4 \text{ Nm}$$ From $\Sigma M_y^P=0$: $$M_{Ay} + M_y + F + A_z = 0 \implies M_{Ay} = -M_y - F - A_z = -4 - 3 - (-3) = -4 - 3 + 3 = -4 \text{ Nm}$$ From $\Sigma M_z^P=0$: $$M_{Az} + M_z + mF + mF_G + A_y = 0 \implies M_{Az} = -M_z - mF - mF_G - A_y = -2 - 3 - 1 - (-4) = -2 - 3 - 1 + 4 = -2 \text{ Nm}$$ 6. **Final answers:** $$A_x = 0 \text{ N}, \quad A_y = -4 \text{ N}, \quad A_z = -3 \text{ N}$$ $$M_{Ax} = -4 \text{ Nm}, \quad M_{Ay} = -4 \text{ Nm}, \quad M_{Az} = -2 \text{ Nm}$$ These reactions balance the applied forces and moments on the rod fixed at the clamp.