# 数学物理 Mathematical Physics MATHS5073_1 /MATHS4107_1

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Now approaching this problem using vector notation, we note that the torque about the fulcrum due to $F_{1}$ is $\tau_{1}=r_{1} \times F_{1}$. Here $r_{1}$ is a vector from the fulcrum to the point of application of $\mathbf{F}{1}$, and $$\tau{1}=\mathbf{r}{1} \times \mathbf{F}{1}=r_{1} F_{1} \sin \theta_{1} \hat{\mathbf{n}}=d_{1} F_{1} \hat{\mathbf{n}}$$
where $\hat{\mathbf{n}}$ is a unit vector normal to, and out of the plane of the figure. A similar analysis of the torque $\tau_{2}$ reveals that it is
$$\tau_{2}=\mathbf{r}{2} \times \mathbf{F}{2}=-r_{2} F_{2} \sin \theta_{2} \hat{\mathbf{n}}=-d_{2} F_{2} \hat{\mathbf{n}}$$
the minus sign occurring because $\hat{\mathbf{n}}$ has the same meaning as before but this torque is directed into the plane of the figure. We now observe that $\theta_{2}=\pi-\theta_{1}$, and therefore $\sin \theta_{2}=\sin \theta_{1}$. At equilibrium the sum of these torques is zero.

## MATHS5073_1 /MATHS4107_1 COURSE NOTES ：

Sometimes the determination of the gradient is straightforward. Given
$$f(x, y, z)=\frac{x y}{z}$$
we easily find
$$\frac{\partial f}{\partial x}=\frac{y}{z}, \quad \frac{\partial f}{\partial y}=\frac{x}{z}, \quad \frac{\partial f}{\partial z}=-\frac{x y}{z^{2}}$$
leading to $\nabla f=\frac{y}{z} \hat{\mathbf{e}}{x}+\frac{x}{z} \hat{\mathbf{e}}{y}-\frac{x y}{z^{2}} \hat{\mathbf{e}}_{z}$