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Some forces are perpendicular to the displacement of a body placed in a "vehicle", if the vehicle at rest. But, the motion of "vehicle" causes the body to move in some direction not perpendicular to the force. Now, the net displacement no longer remains perpendicular to the forces. And they start to do work on the body. We need to find the work done by such forces on the body, in such situations.

To have a feel of what I want to say, Let's have a look at following two problems:-

  1. There is a pendulum in a car, both car and pendulum are at rest. Till now, tension forces appear as if they never do any work on the mass (if displaced slightly). Now, the car starts moving. For the mass to follow string constraint it must also move forwards. The displacement of the mass no longer remains perpendicular to the tension. Hence, tension does work in this situation.

enter image description here

  1. Another such popular example is the work done by Normal force acting on a block placed on a movable inclined wedge, both free to move, over a journey from A to B. Here, Normal does not do any work on the mass(if allowed to slide through a distance), when the wedge is at rest, but the movement of wedge forces normal to do work. For any displacement of the mass relative to the wedge the work done by normal is zero. But, the wedge is also moving.

enter image description here

I just wanted to know whether we could have some nice simplification of the formula $$W_{F} = \int_{A}^{B}\mathbf{F}\cdot\text d\mathbf{s}$$ in order to find the work done by such forces when the "vehicle" is also in motion.

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  • $\begingroup$ There is a pending close review on this question. Offering a bounty stops users from being able to vote to close a question, so I have refunded your bounty. You can offer it again after the close review has concluded. $\endgroup$ – ACuriousMind Aug 6 at 14:39
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    $\begingroup$ I think the reason for the downvote about the question is two fold 1. The head and tail of the question doesn’t really fit together, the “motion” of the vehicle does work (you also forgot the implicit assumption that the “motion” is accelerating motion, if it’s constant velocity, there should be no work done), so what? Of course if the object is accelerating, there should be work. The simplification of the integral F dot ds is another matter altogether! $\endgroup$ – user208685 Aug 12 at 17:17
  • $\begingroup$ 2. The examples you give is also weird, on the first example, yes, the string is doing work, so? If the ball is inside a box, and the box is accelerating, of course something has to accelerate the ball, doing work. On the second example, normal force does not do work! Because on that logic gravity will do work, and it will exactly be in the opposite direction, counteracting the work by the normal force. Or rather, the net force on the block is zero (I’m assuming both block and wedge is moving at constant speed) so the net work is zero as well. $\endgroup$ – user208685 Aug 12 at 17:24
  • $\begingroup$ @user208685 No! even if the vehicle moves with constant velocity there should be work in the both cases see this for the second case (first is similar) physics.stackexchange.com/questions/571554/… . The matter essentially is that there is a component of displacement in the direction of force hence work has to be done. $\endgroup$ – Don't Worry Aug 13 at 2:58
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    $\begingroup$ Take a good look at the comments made by @user208685. He's got a good look on the situation! $\endgroup$ – Deschele Schilder Aug 16 at 8:50
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enter image description here

The work that done by the normal force N is:

$$W=\int N\,sin(\varphi)\,dx=\int N\,sin(\varphi)\,\frac{dx}{dt}\,dt$$

with:

$$N=-m\,\left[\cos(\varphi)\,g+L\dot{\varphi}^2-a\,\sin(\varphi)\right]$$ $$\dot{x}=a\,t$$

you obtain: $$\dot{W}=N\,\sin(\varphi)\,a\,t\tag 1$$

to obtain the result $W(t)$ you need the equation of motion for the pendulum which is:

$$\ddot{\varphi}+\frac{g\,\sin(\varphi)+a\,\cos(\varphi)}{L}=0\tag 2$$

Simulation :

enter image description here

enter image description here

Edit

Normal force

enter image description here

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  • $\begingroup$ @Don'tWorry see new document indeed i forgot the part of the car acceleration $\endgroup$ – Eli Aug 13 at 19:06
  • $\begingroup$ Can you please show, your simulation results? $\endgroup$ – Eli Aug 15 at 5:30
  • $\begingroup$ @Don'tWorry In the accelerating car the pendulum swings also from the left to the right (unless you consider the steady-state). It swings from a point on the left to a lower point on the right (and vice-versa). $\endgroup$ – Deschele Schilder Aug 15 at 18:18
  • $\begingroup$ @descheleschilder Yes! if initially if the mass would be on the right side of the pivot when the car starts accelerating it would again go to the right side. I have assumed that it was initially on the vertical direction. $\endgroup$ – Don't Worry Aug 16 at 3:08
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Using

$$d\vec{s} = d\vec{s_{b/c}} + d\vec{s_{c/g}}$$ where, $d\vec{s}$ is the small displacement of the body wrt ground.
$d\vec{s_{b/c}}$ is small displacement of the mass wrt vehicle.
$d\vec{s_{c/g}}$ is small the displacement of vehicle wrt to the ground.

we can write the work done by such forces as $$W_{F} = \int_{A}^{B}\vec{F}\cdot d\vec{s} = \int_{A}^{B}\vec{F}\cdot (d\vec{s_{b/c}} + d\vec{s_{c/g}}) = \int_{A}^{B}\vec{F}\cdot d\vec{s_{b/c}}+\int_{A}^{B}\vec{F}\cdot d\vec{s_{c/g}}$$ But the such forces relative to the vehicle do not do any work. Hence, the first term vanishes
$$W_{F} = \int_{A}^{B}\vec{F}\cdot d\vec{s_{c/g}}$$

For illustration:
Let's have a look at the problem $1$ stated in the question. We need to find the work done on the mass by tension force in the journey from $A$ to $B$.
Now, the small displacement of the block $d\vec{s}$ can be resolved in two parts, first due to motion of car (say $d\vec{s_c}$) and second due to the circular motion of the mass around the pivot (say $d\vec{r}$). The expression of work done by tension, becomes $$W_{T}=\int_{A}^{B}\vec{T}\cdot d\vec{s} = \int_{A}^{B}\vec{T}\cdot d\vec{s_c}+\int_{A}^{B}\vec{T}\cdot d\vec{r}$$
But, Tension is always perpendicular to $\vec{dr}$. Hence, $$W_{T} = \int_{A}^{B}\vec{T}\cdot d\vec{s_c}$$

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  • $\begingroup$ The work done by the mass against the tension force is always zero, as the mass doesn't move on the string. $\endgroup$ – Deschele Schilder Aug 15 at 5:21
  • $\begingroup$ @descheleschilder Correct! the mass does not move on the string. But, the mass moves forward due to the motion of the car. There is a component of tension along the horizontal direction which accounts for the work. Going on your same approach, what would be the work done by the Normal on the block in the second case? You may say it is zero. But this problem tells that it is not physics.stackexchange.com/questions/571554/… $\endgroup$ – Don't Worry Aug 16 at 3:03
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    $\begingroup$ You state: We need to find the work done on the mass by tension force in the journey from $A$ to $B$. If the pendulum stays vertical (which it does if the vehicle has a constant velocity), no work is done by the tension force. $\endgroup$ – Deschele Schilder Aug 16 at 7:57
  • $\begingroup$ @descheleschilder You say that the pendulum stays vertical. But, that implies that the mass has a velocity in the horizontal equal to the velocity of the vehicle, which it originally didn't have. What do you think would cause the mass to gain velocity in the horizontal direction then? $\endgroup$ – Don't Worry Aug 16 at 9:28
  • $\begingroup$ Why did you think I said the pendulum only differs from hanging vertically if acceleration is present? When the car has reached a velocity, after which it keeps on going with this constant velocity, there's no more acceleration (so no force) anymore in the backward direction and the pendulum would swing as it would while hanging in a gravitational field. $\endgroup$ – Deschele Schilder Aug 16 at 9:36
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In general if the situations are more complex then we have conservation laws and work energy theorem in our pocket.

Both of the situations are not much complex,in $1^{st}$ case:

Working from ground frame we can use work energy theorem:

$$\Delta K=W_{tension}+ W_{gravity}$$

Here the problem shortens easily,because finding the work done by gravity is very easy as it is nothing but negative of change in potential energy of the bob-earth system and change in kinetic energy is already known to us by kinematics.

So the work done by gravity will come out to be equal to: $$W_{gravity}=-mgl(1-\cos\theta)$$ Here $l$ is the length of rope and $\theta$ is a variable angle.

For getting the final speed in $x$ i.e horizontal direction, just resolve the net centripetal acceleration in this direction,then the net acceleration $a_{x}=\frac{dv_{x}}{dt}$ in horizontal direction is given as:

$$\frac{dv_{x}}{dt}=a+\frac{v_{b,c}^{2}\sin\theta}{l}$$

Similarly resolve the centripetal acceleration in $y$ direction,then

$$\frac{dv_{y}}{dt}=\frac{v_{b,c}^{2}\cos\theta}{l}-g$$

Here, $$\vec{v_{b,g}}=v_{x}\hat x+v_{y}\hat y$$ Just solve the above equations and after getting the final speed i.e $v_{b,g}=\sqrt{v_{x}^2+v_{y}^2}$we can get the change in Kinetic energy.

Now work done by tension can be calculated easily

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You state that the car has constant velocity. In the case of the pendulum, the pendulum will remain vertical (or in the same periodic motion). So no extra work will be done by the force of gravity, which is also the case with the mass on a wedge placed on a car with constant velocity.

The situation gets more interesting and less trivial if the cars accelerate (to the right).

The pendulum
The forward acceleration produces a force $-ma$ (the minus sign stand for the acceleration being backward). The total force acting on the mass perpendicular to $r$ is:

$$F_{per}=F_{gper}-F_{aper}=F_g\sin{\theta}-F_a\sin{\theta}$$

This means that the total work, done by the mass, for a half cycle between the points where the velocity of the mass is zero, is:

$$W=\int_{{\theta}_1}^{{\theta}_2}\sin{\theta}(F_g-F_a)d{\theta},$$

which clearly differs (it's smaller) from the stationary case. Of course, the work done in a full cycle is zero (the work done in one direction is opposite to that in the opposite direction), just as in the stationary case.

The mass on the wedge
In the case of the wedge, the force $\vec F$ acting on the wedge by the forward acceleration does cause a change in the work done by the block on the wedge when sliding down the same distance when compared to the non-acceleration case. Obviously the work done gets smaller.

The force acting parallel to the wedge is changed by the acceleration of the wedge to the right. If the acceleration is $\vec a$, the force on the wedge (parallel to the surface of the wedge) will become:

$$ F_{wp}=m {g_p}-m {a_p} ,$$

where $ F_{wp}$ is the total force acting on the mass parallel to the wedge, $m {g_p}$ the force of gravity acting on the mass parallel to the wedge, and $m {a_p}$ the force acting on the mass parallel to the wedge due to the acceleration to the right (which acts to the left, hence the minus sign).
Now, $m {g_p}=m g \cos{\theta}$, where $\theta$ is the angle of inclination,
and $m {a_p}= m a \cos{\theta}$ so,

$$ F_{wp}=(m p-m a)\cos{\theta}.$$

The force of friction is $ {F_f}=\mu{F_n}=\mu F_{wp}\cos{\theta}\sin{\theta}$.

Filling the expression for $ F_{wp}$ into this equation we arrive at the final result:

$${F_f}=\mu{F_n}=\mu(mp-ma){\cos{\theta}}^2\sin{\theta},$$

where $\mu$ is not the $\mu$ in the second image, which is put to zero so the wedge can move without friction on the surface.

So:

$$W_{F} = \int_{A}^{B}F_f ds=\int_{A}^{B}\mu F_nds=\mu(mp-ma){\cos{\theta}}^2\sin{\theta}ds,$$

Note that I didn't use vector notation because in both cases because the integral is done in one direction only. $A$ and $B$ are the endpoints of the distance over which the work is done. Clearly this is less than the work done without acceleration (over the same distance).

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    $\begingroup$ Why do you say the only work done on the body is only done by gravity. Take for example the steady state solution, no oscillation, but accelerating car. The steady state is with the string slightly slanted, but not moving. In this case the motion of the particle in completely horizontal, so the gravity is not doing any work. Still the particle is accelerating with the car, and the string is providing the horizontal force to accelerate the body. $\endgroup$ – A. Jahin Aug 15 at 13:58
  • $\begingroup$ @A.Jahin Obviously, in the steady-state, no work is done! I've worked to the non-steady-state in which the pendulum oscillates and the car accelerates so the force pulling on the mass on the wedge gets less (as can be seen in the formula). $\endgroup$ – Deschele Schilder Aug 15 at 18:09
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Here is a stimulation, I have copied major part of it from here. The only slight difference I have made is change in the effective gravity.

GlowScript 2.1 VPython
top=sphere(pos=vector(0,0,0), radius=0.01)
L=1
th=0
ball=sphere(pos=L*vector(sin(th), -cos(th),0),radius=.05, color=color.red)
string=cylinder(pos=top.pos, axis=(ball.pos-top.pos), radius=0.01, color=color.yellow)
om=0
gg=9.8*sqrt(2);
g=vector(-9.8,-9.8,0)
ball2=sphere(pos=ball.pos, radius=0.05, color=color.cyan)
ball2.m=.05
ball2.p=vector(0,0,0)
attach_trail(ball2)
string2=cylinder(pos=top.pos, axis=(ball2.pos-top.pos), radius=0.01)
t=0
dt=0.001
while t<10:
  rate(1000)
  al = -(gg/L)*sin(th)
  om=om+al*dt
  th=th+om*dt
  ball.pos=L*vector(sin(th),-cos(th),0)
  string.axis=ball.pos-top.pos
  r=top.pos-ball2.pos
  v=mag(ball2.p)/ball2.m
  ball2.a=(v**2)/L
  th2=atan(ball2.pos.x/(-ball2.pos.y))
  T=(ball2.m*gg*cos(th2+pi/4)+ball2.m*ball2.a)*norm(r)
  Fnet=ball2.m*g+T
  ball2.p=ball2.p+Fnet*dt
  ball2.pos=ball2.pos+ball2.p*dt/ball2.m
  string2.axis=ball2.pos-top.pos
  t=t+dt

  
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  • $\begingroup$ I don't see the plot $\varphi(t)$ ? $\endgroup$ – Eli Aug 15 at 9:10
  • $\begingroup$ If you linearized the pendulum differential equation you get: ${\frac {d^{2}}{d{\tau}^{2}}}\varphi \left( \tau \right) +{\frac {g \varphi \left( \tau \right) +a}{L}} =0$ thus the analytical solution ($\varphi(0)=0~,D(\varphi)(0)=0$) is $\varphi(\tau)=\frac{a}{g} \left( \cos \left( {\frac {\sqrt {g}\tau}{\sqrt {L}}} \right) -1 \right) $ this is a SHM with $\omega^2=\frac{g}{L}$ so I think my simulation is correct $\endgroup$ – Eli Aug 15 at 9:25

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