两根绳连接在驳船前部的一个点上。 驳船被水平拉动，沿一条长直运河的中央作直线运动。

一根绳与驳船运动方向成 $25^\circ$ 角，绳中的张力为 $800\,\mathrm{N}$。 如图 2 所示，另一根绳与驳船运动方向成 $\alpha^\circ$ 角，绳中的张力为 $500\,\mathrm{N}$。

两根绳均水平。

(a) 求 $\alpha$ 的值。

驳船的质量为 $15$ 吨，其运动阻力是大小为 $750\,\mathrm{N}$ 的恒力。

(b) 求驳船的加速度。

驳船沿运河直线运动，垂直于航向的两个张力分量必须抵消。原图两绳均朝前，选择反正弦的锐角解。

There is no acceleration perpendicular to the direction of motion, so the transverse tension components balance:

Hence

The acute solution is consistent with both ropes pulling forwards as shown. Retain the unrounded angle for the next part.

先把 15 吨换成 15000 kg。沿航向把两个张力的向前分量相加，再减去阻力，所得合力除以质量。

The mass is $15\,000$ kg. Let the forward acceleration be $a$. Resolving along the direction of motion,

Using the unrounded value of $\alpha$ from (a),

也可先通过力三角形求两张力的合力。三角形中夹角为 $180^\circ-25^\circ-\alpha$，用正弦定理求合力后减阻力。

Let $F$ be the resultant of the two rope tensions; it acts along the canal because their transverse components cancel. In the head-to-tail force triangle, the interior angle opposite $F$ is

The side of length 500 N is opposite $25^\circ$. By the sine rule,

Using the unrounded $\alpha$ from (a), Newton’s second law along the canal gives