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M1 2022 June Q4

A Level / Edexcel / M1

IAL 2022 June Paper · Question 4

题目

Problem

4.

A small block of mass 5 kg5\,\mathrm{kg} lies at rest on a rough horizontal plane.

The coefficient of friction between the block and the plane is 37\frac{3}{7}.

A force of magnitude PP newtons is applied to the block in a direction which makes an angle of 30∘30^\circ with the plane, as shown in Figure 1.

The block is modelled as a particle.

Given that P=14P=14,

(a) find the magnitude of the frictional force exerted on the block by the plane and describe what happens to the block, justifying your answer.

(6)

The value of PP is now changed so that the block is on the point of slipping along the plane.

(b) Find the value of PP.

(6)
题目中文翻译

一个质量为 5 kg5\,\mathrm{kg} 的小物块静止在粗糙水平面上。

物块与平面之间的摩擦系数为 37\frac{3}{7}。

如图 1 所示,一个大小为 PP 牛顿的力作用在物块上,方向与平面成 30∘30^\circ 角。

将物块视为质点。

已知 P=14P=14,

(a) 求平面对物块施加的摩擦力大小,并描述物块会如何运动或保持何种状态,说明答案的理由。

(6)

现改变 PP 的值,使物块恰好要沿平面滑动。

(b) 求 PP 的值。

(6)

解答

(a)

解法一

思路

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先求支持力,再算最大静摩擦力。实际静摩擦力只需平衡水平外力,若所需大小小于最大值,物块仍静止。

答题过程

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Resolving vertically, the normal reaction is

R=5g−14sin⁡30∘=49−7=42 N.R=5g-14\sin30^\circ=49-7=42\ \mathrm N.

The maximum available static friction is

Fmax⁡=μR=37(42)=18 N.F_{\max}=\mu R=\frac37(42)=18\ \mathrm N.

The horizontal component of the applied force is

14cos⁡30∘=73=12.124… N.14\cos30^\circ=7\sqrt3=12.124\ldots\ \mathrm N.

Since 73<187\sqrt3<18, static friction can balance this horizontal force. The block therefore remains at rest, and the actual frictional force has magnitude

F=73 N≈12.1 N\boxed{F=7\sqrt3\ \mathrm N\approx12.1\ \mathrm N}

acting opposite to the horizontal component of the applied force.

(b)

解法一

思路

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临界滑动时才有 F=μRF=\mu R。随着拉力改变,支持力也改变,所以必须重新把支持力表示成 PP 的函数。

答题过程

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Let the new normal reaction be SS. Vertical equilibrium gives

S+Psin⁡30∘=5g,S=49−P2.S+P\sin30^\circ=5g,\qquad S=49-\frac P2.

At limiting equilibrium, horizontal equilibrium gives

Pcos⁡30∘=37S.P\cos30^\circ=\frac37S.

Substituting for SS,

32P= 37(49−P2),73P= 294−3P,P= 29473+3=19.4 N(3 s.f.).\begin{align*} \frac{\sqrt3}{2}P=&\,\frac37\left(49-\frac P2\right),\\ 7\sqrt3P=&\,294-3P,\\ P=&\,\frac{294}{7\sqrt3+3}=\boxed{19.4\ \mathrm N}\quad\text{(3 s.f.)}. \end{align*}

At this value S>0S>0, so contact with the plane is maintained.