Skip to content
CalcGospel 國際數學圖譜
返回

M1 2020 Oct Q4

A Level / Edexcel / M1

IAL 2020 Oct Paper · Question 4

题目

Problem

4.

A non-uniform beam ABAB has length 8 m8\,\mathrm{m} and mass M kgM\,\mathrm{kg}.

The centre of mass of the beam is dd metres from AA.

The beam is supported in equilibrium in a horizontal position by two vertical light ropes. One rope is attached to the beam at CC, where AC=2.5 mAC=2.5\,\mathrm{m} and the other rope is attached to the beam at DD, where DB=2 mDB=2\,\mathrm{m}, as shown in Figure 2.

A gymnast, of mass 64 kg64\,\mathrm{kg}, stands on the beam at the point XX, where AX=1.875 mAX=1.875\,\mathrm{m}, and the beam remains in equilibrium in a horizontal position but is now on the point of tilting about CC.

The gymnast then dismounts from the beam.

A second gymnast, of mass 48 kg48\,\mathrm{kg}, now stands on the beam at the point YY, where YB=0.5 mYB=0.5\,\mathrm{m}, and the beam remains in equilibrium in a horizontal position but is now on the point of tilting about DD.

The beam is modelled as a non-uniform rod and the gymnasts are modelled as particles.

Find the value of MM.

(8)
题目中文翻译

一根非均匀横梁 ABAB 长 8 m8\,\mathrm{m},质量为 M kgM\,\mathrm{kg}。

横梁的质心距 AA 点 dd 米。

两根竖直轻绳使横梁保持水平平衡。如图 2 所示,一根绳连接在 CC 点,其中 AC=2.5 mAC=2.5\,\mathrm{m};另一根连接在 DD 点,其中 DB=2 mDB=2\,\mathrm{m}。

一名质量为 64 kg64\,\mathrm{kg} 的体操运动员站在梁上的 XX 点,其中 AX=1.875 mAX=1.875\,\mathrm{m}。横梁仍保持水平平衡,但此时恰好要绕 CC 点倾斜。

随后,该运动员离开横梁。

第二名质量为 48 kg48\,\mathrm{kg} 的运动员站在梁上的 YY 点,其中 YB=0.5 mYB=0.5\,\mathrm{m}。横梁仍保持水平平衡,但此时恰好要绕 DD 点倾斜。

将横梁视为非均匀杆,将运动员视为质点。

求 MM 的值。

(8)

解答

解法一

思路

展开

每次临界倾覆时,远离转轴的那根绳张力降为零。分别关于临界支点取矩,得到含同一个质心位置的两式;相加后可消去质心位置。

答题过程

展开

The coordinates of the supports measured from AA are 2.5 m2.5\,\mathrm{m} at CC and 6 m6\,\mathrm{m} at DD.

With the 64 kg64\,\mathrm{kg} gymnast at XX, the beam is on the point of tilting about CC, so the tension at DD is zero. Taking moments about CC,

64g(2.5−1.875)=Mg(d−2.5),64g(2.5-1.875)=Mg(d-2.5),

hence

40=M(d−2.5).(1)40=M(d-2.5).\tag{1}

With the 48 kg48\,\mathrm{kg} gymnast at YY, the beam is on the point of tilting about DD, so the tension at CC is zero. Since AY=7.5 mAY=7.5\,\mathrm{m}, taking moments about DD gives

48g(7.5−6)=Mg(6−d),48g(7.5-6)=Mg(6-d),

so

72=M(6−d).(2)72=M(6-d).\tag{2}

Adding (1) and (2),

112=M((d−2.5)+(6−d))=3.5M.112=M\big((d-2.5)+(6-d)\big)=3.5M.

Therefore M=32\boxed{M=32}. The beam has mass 32 kg32\,\mathrm{kg}. Substitution into (1) gives d=3.75 md=3.75\,\mathrm{m}, between CC and DD as required by both limiting positions.