讲解这道题---**Question:**
2. (14 分) 小明同学平时注意锻炼身体,力量较大,最多能提起 m=50kg 的物体。一重物放置在倾角 θ=15°的粗糙斜坡上,重物与斜坡间的摩擦因数为 μ = \frac{\sqrt{3}}{3} \approx 0.58。试求该同学向上拉动的重物质量 M 的最大值?
**Given Information:**
* Maximum mass the student can lift vertically: m = 50 kg.
* Inclination angle of the rough slope: θ = 15°.
* Coefficient of friction between the object and the slope: μ = \frac{\sqrt{3}}{3} \approx 0.58.
* Mathematical Formula: (已知 a cosθ + b sinθ = \sqrt{a^2 + b^2} sin(α + θ), sin α = \frac{a}{\sqrt{a^2 + b^2}})
**Diagram Description:**
* **Type:** Diagram illustrating an object on an inclined plane with an applied force.
* **Main Elements:**
* An inclined plane making an angle θ with the horizontal ground.
* A block labeled M placed on the inclined plane.
* A force vector labeled F originating from the block, directed upwards along the inclined plane, appearing slightly above the plane. An arrowhead indicates the direction of the force.
* An angle θ is labeled between the inclined plane and the horizontal line representing the ground.
* The horizontal ground is indicated by a jagged line pattern.
**Notes:**
* The problem context implies that the maximum upward pulling force the student can exert is equal to the force required to lift 50kg vertically, i.e., F_max = m*g = 50g (where g is the acceleration due to gravity).
* The force F in the diagram represents the pulling force exerted by the student.
* The question asks for the maximum mass M that can be pulled *upwards* along the slope.
视频信息
答案文本
视频字幕
这是一道关于斜面拉力的物理问题。小明同学最多能提起50千克的物体，现在要在倾角15度、摩擦因数为根号3除以3的粗糙斜面上向上拉动重物。我们需要分析受力情况，求出能拉动的最大质量。
首先进行受力分析。重物M在斜面上受到四个力的作用：重力Mg竖直向下，支持力N垂直于斜面向上，摩擦力f沿斜面向下阻碍运动，以及小明施加的拉力F沿斜面向上。这四个力的平衡关系决定了物体的运动状态。
接下来建立力的平衡方程。将重力分解为垂直于斜面和沿斜面的两个分量。垂直于斜面方向，支持力N等于重力的垂直分量Mg余弦θ。沿斜面方向，拉力F需要克服重力的平行分量Mg正弦θ和摩擦力f。摩擦力等于μ乘以支持力N。最终得到拉力F等于Mg乘以正弦θ加μ余弦θ。
现在进行三角函数的计算。已知倾角θ等于15度，摩擦因数μ等于根号3除以3。我们需要计算sin15度和cos15度的值。利用三角恒等式，sin15度等于sin括号45度减30度，展开后得到根号6减根号2，全部除以4。同样地，cos15度等于cos括号45度减30度，展开后得到根号6加根号2，全部除以4。
现在进行最终计算。小明的最大拉力等于50g，根据前面推导的公式，最大质量等于50除以括号sin θ加μ cos θ。将计算出的三角函数值代入，经过化简得到分母为根号6除以3。因此最大质量等于50乘以3除以根号6，等于150除以根号6。有理化分母后得到25倍根号6千克。所以该同学能拉动的最大质量为25根号6千克。