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M2 2022 June Q7

A Level / Edexcel / M2

IAL 2022 June Paper · Question 7

题目

Problem

7.

The template shown in Figure 3 is formed by joining together three separate laminas. All three laminas lie in the same plane.

  • PQUVPQUV is a uniform square lamina with sides of length 3a3a
  • URSTURST is a uniform square lamina with sides of length 6a6a
  • QRUQRU is a uniform triangular lamina with UQ=3aUQ=3a, UR=6aUR=6a and angle QUR=90∘QUR=90^\circ

The mass per unit area of PQUVPQUV is kk, where kk is a constant. The mass per unit area of URSTURST is kk. The mass per unit area of QRUQRU is 2k2k.

The distance of the centre of mass of the template from QTQT is dd.

(a) Show that d=2914ad=\frac{29}{14}a

(5)

The template is freely suspended from the point QQ and hangs in equilibrium with QRQR at θ∘\theta^\circ to the downward vertical.

(b) Find the value of θ\theta

(7)
题目中文翻译

图 3 所示的模板由三块独立薄片连接而成。 三块薄片均位于同一平面内。

  • PQUVPQUV 是边长为 3a3a 的均匀正方形薄片;
  • URSTURST 是边长为 6a6a 的均匀正方形薄片;
  • QRUQRU 是均匀三角形薄片,其中 UQ=3aUQ=3a,UR=6aUR=6a,且 ∠QUR=90∘\angle QUR=90^\circ。

PQUVPQUV 的单位面积质量为 kk,其中 kk 为常数。 URSTURST 的单位面积质量为 kk。 QRUQRU 的单位面积质量为 2k2k。

模板质心到 QTQT 的距离为 dd。

(a) 证明 d=2914ad=\frac{29}{14}a。

(5)

模板从 QQ 点自由悬挂并处于平衡状态,QRQR 与竖直向下方向成 θ∘\theta^\circ 角。

(b) 求 θ\theta 的值。

(7)

解答

(a)

解法一

思路

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三角形面积只有大正方形的四分之一,但其面密度加倍。先求真实质量比,再以 QTQT 为轴取矩,小正方形在轴左侧要用负距离。

答题过程

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The masses of the two squares and the triangle are

m1=k(3a)2=9ka2,m_1=k(3a)^2=9ka^2, m2=k(6a)2=36ka2,m_2=k(6a)^2=36ka^2, m3=2k(12×3a×6a)=18ka2.m_3=2k\left(\frac12\times3a\times6a\right)=18ka^2.

Measure horizontal distances to the right from QTQT. The centroid distances of PQUVPQUV, URSTURST and QRUQRU are respectively

−3a2,3a,6a3=2a.-\frac{3a}{2},\qquad3a,\qquad\frac{6a}{3}=2a.

Taking moments about QTQT and cancelling ka2ka^2,

63d=9(−3a2)+36(3a)+18(2a).63d=9\left(-\frac{3a}{2}\right)+36(3a)+18(2a).

Thus

63d=(−272+108+36)a=2612a,63d=\left(-\frac{27}{2}+108+36\right)a=\frac{261}{2}a,

and therefore

d=2914a.\boxed{d=\frac{29}{14}a}.

(b)

解法一

思路

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还须求质心在原图中距 PQPQ 向下的距离。悬挂时 QGQG 竖直向下,所求角是 QRQR 与 QGQG 的夹角;两者在原图中相对于 QTQT 的角度相减即可。

答题过程

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Let GG be the centre of mass. Measure vertical distances downwards from the horizontal line PQPQ, and let the centroid distance be zz.

The centroid distances for PQUVPQUV, URSTURST and QRUQRU are respectively

3a2,3a+3a=6a,2a.\frac{3a}{2},\qquad3a+3a=6a,\qquad2a.

Taking moments about PQPQ,

63z=9(3a2)+36(6a)+18(2a).63z=9\left(\frac{3a}{2}\right)+36(6a)+18(2a).

Hence

z=59a14.z=\frac{59a}{14}.

The angle ϕ\phi between QTQT and QGQG therefore satisfies

tan⁡ϕ=dz=2959.\tan\phi=\frac dz=\frac{29}{59}.

In right-angled triangle QURQUR, the angle γ\gamma between QTQT and QRQR satisfies

tan⁡γ=URQU=2.\tan\gamma=\frac{UR}{QU}=2.

When the template is suspended from QQ, QGQG is vertically downwards.

Since QGQG is between QTQT and QRQR,

θ=γ−ϕ=tan⁡−12−tan⁡−12959.\theta=\gamma-\phi=\tan^{-1}2-\tan^{-1}\frac{29}{59}.

Thus

θ=37.3∘.\boxed{\theta=37.3^\circ}.