站在高一学生的角度解答，用中文讲解，要用图形结合解答。---**Question 13**
**Question Stem:**
A wooden wedge of mass M and inclination angle θ remains at rest on a horizontal surface. When a wooden block of mass m is placed on the inclined surface of the wedge, it slides down uniformly. If a force F, which makes an angle α with the inclined surface of the wooden wedge, pulls the wooden block to slide up uniformly along the inclined surface, as shown in the figure, then:
**(1)** When α is how much, the pulling force F has a minimum value? Find this minimum value.
**(2)** When α=0, what is the friction force exerted by the wooden wedge on the horizontal surface?
**Chart/Diagram Description:**
* **Type:** A physics diagram illustrating a block on an inclined plane.
* **Main Elements:**
* **Wedge (Inclined Plane):** A large right-angled triangular block is shown resting on a horizontal flat surface. It is labeled with mass "M". The angle of inclination of its sloped surface with the horizontal ground is denoted by "θ".
* **Block:** A smaller rectangular block, labeled with mass "m", is placed on the inclined surface of the wedge.
* **Force F:** An arrow originates from the block "m" and points upwards and slightly away from the inclined surface. This arrow represents a force, labeled "F".
* **Angle α:** A dashed line extends along the inclined surface from the position of block "m". The angle between this dashed line (representing the direction parallel to the incline) and the force vector "F" is labeled "α".
* **Horizontal Surface:** The flat ground beneath the wedge.
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我们来分析一个斜面力学问题。质量为M的楔形体静置在水平面上，倾斜角为θ。当质量为m的木块放在斜面上时，它会匀速下滑。现在我们用一个与斜面成α角的拉力F，使木块匀速上滑。我们需要分析这个系统的力学特性。
首先分析木块匀速下滑时的受力情况。木块受到三个力：重力mg向下，正压力N垂直斜面向上，摩擦力f沿斜面向上。将重力分解为平行斜面和垂直斜面的分量。由于木块匀速运动，各方向受力平衡。平行斜面方向：mg sin θ等于摩擦力f。垂直斜面方向：mg cos θ等于正压力N。根据摩擦定律f等于μN，可得动摩擦系数μ等于tan θ。这是解决问题的关键参数。
现在分析施加拉力F后木块匀速上滑的受力情况。木块受到四个力：重力mg向下，正压力N垂直斜面向上，摩擦力f沿斜面向下，拉力F与斜面成α角向上。将拉力F分解为两个分量：平行斜面的分量F cos α和垂直斜面的分量F sin α。由于木块匀速运动，建立两个方向的平衡方程。平行斜面方向：F cos α等于mg sin θ加上摩擦力f。垂直斜面方向：正压力N加上F sin α等于mg cos θ。
现在求解拉力F的最小值。由垂直方向平衡方程可得正压力N，再由摩擦定律得到摩擦力f。将这些代入平行方向平衡方程，整理后得到拉力F的表达式。F等于mg sin θ乘以1加μ，除以cos α加μ sin α。要使F最小，需要分母最大。对分母求导并令其为零，得到tan α等于μ，即tan α等于tan θ，所以α等于θ时拉力最小。将α等于θ代入，得到最小拉力F min等于mg sin 2θ除以2。
现在分析第二个问题：当α等于0时，拉力F水平向上。此时拉力大小为mg sin θ乘以1加μ，即2mg sin θ cos θ。为了求地面对楔形体的摩擦力，我们分析整个楔形体和木块组成的系统。在水平方向，系统受到拉力F向右和地面摩擦力向左。由于系统保持静止，水平方向力平衡，所以地面对楔形体的摩擦力等于拉力F，即mg sin 2θ。