摩擦公式的问题

摩擦力是日常生活中几乎无处不在的一种力量。这种力使我们能够行走、握住物体等。简而言之，它对于现实生活中几乎所有的身体动作都是必不可少的。当一个球以一定的速度扔到地板上时，它会继续移动一段时间，然后停止。在理想情况下，球应该保持移动，因为没有外力作用在球上。但它会静止，这意味着存在一种抵抗运动的力。这种力称为摩擦力。摩擦力本质上反对物体之间的各种运动。

摩擦

摩擦力与运动相反。虽然它通常会产生热量并磨损物体，但它通常很有帮助。它提供了行走和跑步所需的必要牵引力而不会打滑。在汽车中，它不太受欢迎，因为它占能量损失的 20%。摩擦力取决于作用在表面上的法向力。摩擦力可以分为两部分：

1. 静摩擦：这是一种防止物体在表面上滑动的摩擦。该方向阻止对象相互滑动并且平行于表面。
2. 动摩擦：当物体在表面上滑动时，这种类型的摩擦会起作用。方向与静摩擦力相同。

摩擦定律：

1. Friction of the moving object is proportional and perpendicular to the normal force applied on it.
2. The friction experienced by the object depends on the nature of the surface it is in contact with.
3. Friction is independent of the area of contact as long as there is an area of contact.
4. Kinetic friction is independent of velocity.
5. The coefficient of static friction is greater than the coefficient of kinetic friction.

摩擦方程

当对物体施加力时，假设 F _N是正常作用在表面上的力。静摩擦值由 F _s和 $\mu_s$ 是静摩擦系数的值。那么控制静摩擦力的方程由下式给出，

$F_s \leq \mu_sF_N$

在物体运动的情况下，摩擦力将由下式给出，

$F_k = \mu_kF_N$

摩擦系数由下式给出，

$\frac{F}{N} = \mu$

这里，F 是摩擦力，N 是法向力的值。

让我们看看与这些概念相关的一些问题。

示例问题

问题 1：静摩擦系数为 0.4 的盒子上施加了 60N 的力。求摩擦力。

回答：

Force of static friction is given by, 

$F_s = \mu_sF_N$

Here, $\mu_s$ is the coefficient of static friction and N is the normal force. 

 $\mu_s = 0.4$ and N = 60N 

Plugging the values in the equation, 

$F_s = \mu_sF_N \\ = F_s = (0.4)(60) \\ = F_s = 24 \text{ N}$

问题 2： 100N 的力施加在静摩擦系数为 0.2 的地板上的盒子上。找到摩擦力

回答：

Force of static friction is given by, 

$F_s = \mu_sF_N$

Here, $\mu_s$ is the coefficient of static friction and N is the normal force. 

 $\mu_s = 0.2$ and N = 100N 

Plugging the values in the equation, 

$F_s = \mu_sF_N \\ = F_s = (0.2)(100) \\ = F_s = 20 \text{ N}$

问题3：在静摩擦系数为0.2的箱子上以30°角施加100N的力。找到摩擦力

回答：

Force of static friction is given by, 

$F_s = \mu_sF_N$

Here, $\mu_s$ is the coefficient of static friction and N is the normal force. 

 $\mu_s$ = 0.2, $\theta$ = 30° and F = 100N

N = Fsin($\theta$) 

⇒N = (100)sin(30°) 

⇒N = 50

Plugging the values in the equation, 

$F_s = \mu_sF_N \\ = F_s = (0.2)(50) \\ = F_s = 10 \text{ N}$

问题4：200N 的力以30°角施加在静摩擦系数为0.2 的地板上的盒子上。找到摩擦力

回答：

Force of static friction is given by, 

$F_s = \mu_sF_N$

Here, $\mu_s$ is the coefficient of static friction and N is the normal force. 

 $\mu_s$ = 0.2, $\theta$ = 30° and F = 200N

N = Fsin($\theta$) 

⇒N = (200)sin(30°) 

⇒N = 100

Plugging the values in the equation, 

$F_s = \mu_sF_N \\ = F_s = (0.2)(100) \\ = F_s = 20 \text{ N}$

问题5：在静摩擦系数为0.2的40Kg箱子上水平施加60N的力。求摩擦力。

回答：

Force of static friction is given by, 

$F_s = \mu_sF_N$

Here, $\mu_s$ is the coefficient of static friction and N is the normal force. 

 $\mu_s = 0.2$ and N = mg

N = mg 

⇒N = (40)(10)

⇒N = 400N 

Static friction will be, 

$F_s = \mu_sF_N \\ F_s = (0.2)(400) \\ F_s = 80 \text{ N}$

Now, the applied force is 60N and the maximum static friction is 80N. Since the applied force is less than the value of static friction, the box will not move and the frictional force will be equal to the applied force. 

问题6：50N 的力水平施加在静摩擦系数为0.3 的静摩擦系数为50Kg 的箱子上。求摩擦力。

回答：

Force of static friction is given by, 

$F_s = \mu_sF_N$

Here, $\mu_s$ is the coefficient of static friction and N is the normal force. 

 $\mu_s = 0.3$ and N = mg

N = mg 

⇒N = (50)(10)

⇒N = 500N 

Static friction will be, 

$F_s = \mu_sF_N \\ F_s = (0.3)(500) \\ F_s = 150 \text{ N}$

Now, the applied force is 100N and the maximum static friction is 150N. Since the applied force is less than the value of static friction, the box will not move and the frictional force will be equal to the applied force. 

问题7：在静摩擦系数为0.4，动摩擦系数为0.2的地板上放置一个2Kg的箱子，施加40N的水平力。求摩擦力。

回答：

Force of static friction is given by, 

$F_s = \mu_sF_N$

Here, $\mu_s$ is the coefficient of static friction and N is the normal force. 

 $\mu_s = 0.4$ and N = mg 

N = mg 

⇒N = (2)(10) 

⇒N = 20N 

Plugging the values in the equation, 

$F_s = \mu_sF_N \\ = F_s = (0.4)(20) \\ = F_s = 8 \text{ N}$

Now, since the maximum static friction is less than the force applied. The box will start moving and then the kinetic friction will act on the body.

$F_k = \mu_kF_N\\ = F_k = (0.2)(mg) \\ = F_k = (0.2)(2 \times 10) \\ = F_k = 4 \text{ N}$

问题8：在静摩擦系数为0.4，动摩擦系数为0.2的地板上放置一个3Kg的箱子，施加50N的水平力。求摩擦力。

回答：

Force of static friction is given by, 

$F_s = \mu_sF_N$

Here, $\mu_s$ is the coefficient of static friction and N is the normal force. 

 $\mu_s = 0.4$ and N = mg 

N = mg 

⇒N = (3)(10) 

⇒N = 30N 

Plugging the values in the equation, 

$F_s = \mu_sF_N \\ = F_s = (0.4)(30) \\ = F_s = 12 \text{ N}$

Now, since the maximum static friction is less than the force applied. The box will start moving and then the kinetic friction will act on the body.

$F_k = \mu_kF_N\\ = F_k = (0.2)(mg) \\ = F_k = (0.2)(3 \times 10) \\ = F_k = 6 \text{ N}$