解决力学问题
到现在为止,人们可能已经听说过牛顿运动定律。这些法律将帮助您解决机械问题。通常,力学问题不包括作用于单个项目的大量力。相反,它与许多物体的集合有关,除了感受引力外,它们还相互施加力。在这篇文章中,我们将研究各种机械问题解决策略。
在尝试回答力学问题时,请记住您可以选择组件的任何部分并将运动定律应用于该部分。您需要做的就是考虑由于装配的剩余部分而作用在“选定零件”上的所有力。为简单起见,我们将装配的选定组件称为“系统”,其余部分称为“环境”。
牛顿第一运动定律
该定律也称为惯性定律。如果物体上的净外力为零,则其加速度为零。只有当身体上有净外力时,加速度才能非零。
Σ F = 0
⇒ dv/dt = 0
其中,F 是力(F 的总和表示施加的净力),v 是物体的速度。
牛顿第一运动定律的应用:
- 一个物体被抛到外太空,以零加速度沿同一方向移动,直到任何其他外部物体以某种力撞击它为止。
- 只要没有净力作用,放在桌子上的一本书就会保持静止。
- 由于惯性,马拉松运动员继续跑出终点线几米。
牛顿第二运动定律
该定律也称为动量定律。物体动量的变化率与所施加的力成正比,并且发生在力作用的方向上。
F = dp/dt
其中,dp 是动量随时间 dt 变化的变化。
牛顿第二运动定律的应用:
- 在超市里推一个空的推车比推一个满载的推车更容易。更大的质量需要更多的动力来加速。
- 物体从一定高度落下,由于所施加的重力,加速度会增加。
牛顿第三运动定律
该定律也称为作用定律和反应定律。每当一个物体对另一个物体施加力时,第二个物体对第一个物体施加相等且相反的力。
F A = -F B
F 12 = F 21
牛顿第三运动定律的应用:
- 当我们拉动松紧带时,它会自动回到原来的位置。动作(施加的力)以能量的形式存储,并以相等且相反的力的反作用力释放。
- 当火箭发射时,燃烧气体排出的力(作用)对火箭施加相等且相反的力(反应)并向上移动。
惯性
In practical, the particle does not change its state of rest or of uniform motion along a straight line unless it is forced to do this. This tendency of of particle to do not change its state of rest or state of uniform motion along a straight line, unless that state is changed by an external force is called as inertia.
质量是完全取决于物体惯性的量。物体的惯性越大,它的质量就越大。粒子的质量越大,加速度越小,因此惯性越大。
摩擦
反对物体在另一个物体表面上的相对运动的性质称为摩擦力。
F = μ N
其中 μ 是摩擦系数,N 是法向力。
例如
- 行走时地面和鞋子之间的摩擦力可以防止我们滑倒。
- 如果没有摩擦,从电机到机器的皮带无法覆盖运动。
在通过任何与问题相关的牛顿运动定律之前。你必须对与之相关的所有概念有很强的掌控力。物理学是一门帮助我们了解世界的学科。你应该学习物理学,因为你正在帮助自己了解世界上不同的现象是如何发生的。牛顿运动定律最核心的秘密是自由体图(FBD),它可以帮助您非常轻松地解决问题。
示例问题
问题1:一名乘客坐在时速100公里的火车上打电话时,不小心将手机从窗户上掉了下来。不计空气摩擦,手机刚落地前的水平速度是多少?
回答:
According to Newton’s first law of motion, object in a motion tends to stay in a motion unless until any external force is not acting. As there is no air friction acting on a object (mobile phone) to slow down the object in the horizontal direction after it drops from the train and acceleration due to gravity would only affect in the vertical direction. So, horizontal speed of the mobile phone just before hitting ground would be approximately 100 km/hr.
问题 2:保持 1.5 kg 的球以 40 m/s 的恒定速度运动需要多大的净力?
回答:
According to Newton’s first law of motion,every body continues to be in its state of rest or of uniform motion in a straight line until unless any external force is not acting.
If the net external force on a body is zero, its acceleration is zero.Hence force needed is also zero. Therefore 0 N net force required to to keep ball moving with constant velocity of 40 m/s.
问题3:一艘2000公斤的宇宙飞船以1200米/秒的恒定速度在太空中运动。什么是作用在飞船上的净力(没有重力作用在飞船上)。
回答:
Newton’s first law of motion states that object remains in a motion until unless any external force is not acting on a object.In a space there is vacuum and there is no external air resistance.Hence, spaceship will travel at constant velocity of 1200 m/s with zero acceleration.
Since, m= mass of spaceship = 2000 kg
a= acceleration of spaceship = 0
∑F = m×a
= 2000 × 0
= 0 N
Hence, net force is acting on a spaceship is 0 N.
问题4:静摩擦和动摩擦是什么意思?
回答:
Resistance encountered by a body in static condition while tending to move under the action of an external force is called static friction. In static friction, the frictional force resists force that is applied to an object, and the object remains at rest until the force of static friction is overcome.It is denoted as μs.
The resistance encountered by sliding body on a surface is known as kinetic friction. Kinetic friction is denoted as μk. Kinetic friction is defined as a force that acts between moving surfaces. A body moving on the surface experiences a force in the opposite direction of its movement. The magnitude of the force will depend on the coefficient of kinetic friction between the two materials.
问题 5:如果一辆质量为 200 kg 的汽车以 5 m/s 2的加速度运动,那么汽车的净力是多少?
回答:
Given that,
Mass of a car = Mc = 200 kg
Acceleration of a car = ac =5 m/s2
Using formula F = Mc × a
= 200 × 5
= 1000 N
Therefore, the net Force is 1000 N.
问题6:击球手以20m/s的速度向投球手的方向直击球,球的初速度为12m/s。假设球的质量为0.10kg。确定球上动量的变化。
回答 :
Given that,
Initial velocity of the ball = 12 m/s
Final velocity of the ball = 20 m/s
Mass of the ball = 0.10kg
Change in momentum = final momentum – initial momentum
= m×v2 – m×v1
= 0.10×20 – (-0.10×12) (ball again is in the direction from the batsman to the bowler)
= 3.2 N.s
Therefore, the change in momentum is 3.2 N.s.
问题 7:在训练期间,一名警察用枪向木块发射了一颗子弹,现在一颗质量为 10 gm 的子弹以 400 m/s 的速度移动,在停止前穿透木块 4 cm。假设木块施加的力是均匀的,求力的大小?
解决方案:
Given that,
Mass of the bullet = Mb = 10 gm = 0.010 kg
Penetration of bullet before coming to rest = s = 4 cm = 0.04 m.
Initial velocity of bullet = Vi =400 m/s
Final velocity of bullet = Vf = 0 m/s
Here, wooden block will exert force opposite in the direction of velocity,therefore this force causes deceleration. Hence a be the deceleration in this case (-a)
By using kinematic equation,
(Vf)2 = (Vi)2 + 2as ——(1)
0 = (400)2 – 2 × a × 0.04
a = ( (400)2 – 0 ) / 2 × 0.04
= 160000 / 0.08
= 2000000
The force on the bullet = Mb × a
= 0.01 × 2000000
= 20000 N
问题 8:将一个质量为 100 公斤的盒子放在地板上,对地板施加一些力。确定地板对箱子施加的力是多少? (这里 g= 9.81 m/s 2 )。
回答:
According to Newton’s third law motion,every action there is equal and opposite reaction.Hence the force exerted by the floor on the box will be the weight of box.
Given that,
Mass of box = M = 100 kg.
weight of the box = M × g
= 100 × 9.81
= 981 N
The force exerted by the floor on the box = -981 N
This Negative sign indicates that force applied by floor is in opposite direction of force applied by the box.
Therefore, the Force applied by the floor is equal to 981 N.
问题 9:定义静止、运动和方向的惯性?
回答:
A characteristic of matter that allows it to remain in its current condition of rest or uniform motion in a straight line until it is disrupted by an external force is called an inertia.
- Inertia of rest: The inability of a body to change its state of rest by itself is called inertia of rest
- Inertia of motion: The inability of a body to change its state of motion by itself is inertia of motion.
- Inertia of direction: The inability of a body to change its direction of motion by itself inertia of direction.
问题 10:电梯中有两名乘客,他们的质量向下施加 180 N 的力。他们从 207 N 的电梯楼层受到一个向上的法向力。他们以什么速度向上加速? (这里g=10 m/s 2 )
回答:
Given that,
Upward force = 207 N
Downward force = 180 N
As they are accelerating in the upward direction then the net force-
Net Force = ∑F = Upward force – Downward force
= 207 -180
= 27 N
To find total mass of the passengers, use the equation for force of gravity,
F = m×g
m = 180/10
m = 18 Kg
To find net acceleration, use Newton’s second law of motion,
F = m × a
a = 27/18
a = 1.5 m/s2
Therefore, they are accelerating in the upward direction at the rate of 1.5m/s2.