题目

Problem

A car of mass $800\,\mathrm{kg}$ is towing a trailer of mass $400\,\mathrm{kg}$ up a straight road using a towbar. The towbar is parallel to the road and parallel to the direction of motion of the car.

The road is inclined to the horizontal at an angle $\alpha$ , where $\sin\alpha=\frac{1}{7}$ . The engine of the car produces a constant driving force of magnitude $D$ newtons. The resistance to the motion of the car from non-gravitational forces is modelled as a single force of magnitude $420\,\mathrm{N}$ . The resistance to the motion of the trailer from non-gravitational forces is modelled as a single force of magnitude $300\,\mathrm{N}$ . The car and trailer are modelled as particles and the towbar is modelled as a light rod.

Given that the tension in the towbar is $2060\,\mathrm{N}$ , find the value of $D$ .

(7)

题目中文翻译

一辆质量为 $800\,\mathrm{kg}$ 的汽车通过牵引杆，沿一条笔直的上坡公路拖动质量为 $400\,\mathrm{kg}$ 的拖车。牵引杆平行于公路及汽车的运动方向。

公路与水平面成 $\alpha$ 角，其中 $\sin\alpha=\frac{1}{7}$ 。汽车发动机产生大小为 $D$ 牛顿的恒定驱动力。将汽车所受的非重力运动阻力视为大小为 $420\,\mathrm{N}$ 的单个力，将拖车所受的非重力运动阻力视为大小为 $300\,\mathrm{N}$ 的单个力。将汽车和拖车视为质点，将牵引杆视为轻杆。

已知牵引杆中的张力为 $2060\,\mathrm{N}$ ，求 $D$ 的值。

(7)

解答

解法一

思路

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先单独研究拖车，由已知牵引力求共同加速度；再研究汽车，牵引杆对汽车的力方向向后。

答题过程

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Take up the slope as positive, and let the common acceleration be $a$ . For the trailer,

2060-300-400g\sin\alpha=400a.

Using $g=9.8$ and $\sin\alpha=1/7$ ,

2060-300-560=400a, \qquad a=3.0\,\mathrm{m\,s^{-2}}.

For the car, the towbar force acts down the slope, so

D-420-800g\sin\alpha-2060=800a.

Thus

D-420-1120-2060=2400, \qquad \boxed{D=6.00\times10^3\,\mathrm{N}}.

解法二

思路

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求出拖车加速度后，也可以用整个系统的运动方程求驱动力，内部牵引力会抵消。

答题过程

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The trailer equation, as in method 1, gives $a=3.0\,\mathrm{m\,s^{-2}}$ . For the car and trailer together,

D-(420+300)-(800+400)g\sin\alpha=(800+400)a.

Substitution gives

D-720-1680=1200(3), \qquad \boxed{D=6.00\times10^3\,\mathrm{N}}.

M1 2019 Oct Q3

题目

解答

解法一

思路

答题过程

解法二

思路

答题过程