题目

Problem

A van of mass $900\,\mathrm{kg}$ is moving along a straight horizontal road. The van is towing a trailer of mass $300\,\mathrm{kg}$ . The trailer is attached to the van by a rigid towbar which is parallel to the direction of motion of the van and the trailer, as shown in Figure 2.

The resistance to the motion of the van is modelled as a constant force of magnitude $500\,\mathrm{N}$ . The resistance to the motion of the trailer is modelled as a constant force of magnitude $200\,\mathrm{N}$ . The towbar is modelled as a light rod.

The engine of the van is working at a constant rate of $18\,\mathrm{kW}$ .

At the instant when the speed of the van is $12\,\mathrm{m\,s^{-1}}$ , the tension in the towbar has magnitude $T$ newtons.

(a) Find the value of $T$

(7)

Later, the van is towing the trailer up a straight road inclined at an angle $\alpha$ to the horizontal, where $\sin\alpha=\frac1{15}$ The towbar is still parallel to the direction of motion of the van and the trailer.

The resistance to the motion of the van from non-gravitational forces is modelled as a constant force of magnitude $500\,\mathrm{N}$ . The resistance to the motion of the trailer from non-gravitational forces is modelled as a constant force of magnitude $200\,\mathrm{N}$ . The towbar is again modelled as a light rod.

When the van is travelling with a constant speed of $v\,\mathrm{m\,s^{-1}}$ , the engine of the van is working at a constant rate of $18\,\mathrm{kW}$ .

(b) Find the value of $v$

(4)

题目中文翻译

质量为 $900\,\mathrm{kg}$ 的厢式货车沿水平直路行驶。货车牵引质量为 $300\,\mathrm{kg}$ 的拖车。如图 2 所示，拖车通过刚性牵引杆与货车连接，牵引杆平行于货车与拖车的运动方向。

将货车所受的运动阻力视为大小为 $500\,\mathrm{N}$ 的恒力。将拖车所受的运动阻力视为大小为 $200\,\mathrm{N}$ 的恒力。将牵引杆视为轻杆。

货车发动机以恒定功率 $18\,\mathrm{kW}$ 工作。

在货车速率为 $12\,\mathrm{m\,s^{-1}}$ 的瞬间，牵引杆中的张力大小为 $T$ 牛顿。

(a) 求 $T$ 的值。

(7)

后来，货车牵引拖车沿一条与水平面成 $\alpha$ 角的直路上坡，其中 $\sin\alpha=\frac1{15}$ 。牵引杆仍平行于货车与拖车的运动方向。

将货车所受的非重力运动阻力视为大小为 $500\,\mathrm{N}$ 的恒力。将拖车所受的非重力运动阻力视为大小为 $200\,\mathrm{N}$ 的恒力。仍将牵引杆视为轻杆。

当货车以恒定速率 $v\,\mathrm{m\,s^{-1}}$ 行驶时，发动机以恒定功率 $18\,\mathrm{kW}$ 工作。

(b) 求 $v$ 的值。

(4)

解答

(a)

解法一

思路

展开

水平路段也未必匀速。先用功率求牵引力，再以整体求加速度，最后单独研究拖车求杆张力。

答题过程

展开

The engine power is 18000 W. At speed 12 m s $^{-1}$ , its driving force is

F=\frac{18000}{12}=1500\text{ N}.

For the van and trailer together, resolving forwards,

1500-(500+200)=(900+300)a.

Hence

a=\frac{800}{1200}=\frac23\text{ m s}^{-2}.

For the trailer alone,

T-200=300a.

Therefore

T=200+300\left(\frac23\right),

\boxed{T=400\text{ N}}.

(b)

解法一

思路

展开

本问明确恒速，整体沿坡合力为零。功率仍是 $18000$ 瓦，但牵引力须平衡两项阻力以及整体重力沿坡分量。

答题过程

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At constant speed the acceleration is zero. Resolving up the road for the combined system,

F-(500+200)-(900+300)g\sin\alpha=0.

Using $F=18000/v$ and $\sin\alpha=1/15$ ,

\frac{18000}{v}=700+1200(9.8)\left(\frac1{15}\right).

Thus

\frac{18000}{v}=1484,

and hence

\boxed{v=12.1\text{ m s}^{-1}}.

M2 2025 Jan Q2

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程