86

Absolutely you can apply a force on some other object and make it accelerate. What will happen is that you too will accelerate - in the opposite direction. You push the thing, the thing pushes you, it's classic Newton's Third Law. This is akin to rocket propulsion.


21

You can apply a force to the object, but since you are on a frictionless surface, you will slide backwards. In your example of standing on a frictionless surface and pushing another object on that same surface, you exert a force $F$ on the object and the object exerts a force $F$ of equal magnitude and opposite direction on you, according to Newton's third ...


9

To add to the other answers: The reason you ask this question might be that you intuitively and from other physics teaching know that you can't move anywhere on a frictionless surface. That you will never be able to set off and move away from your starting point. This is true. Still, though, while you can't move as a whole, you can move your arms and legs ...


5

The string is terminated by a small loop around the axle. The loop is just a bit larger than the diameter of the axle so that, when the yoyo is fully unwound and spinning rapidly, the yoyo can continue spinning with its axle sliding on the lower part of the small loop. The tension due to the weight of the yoyo stops the string being swept around the axle. ...


4

If $F=f_{friction}$ then, as you say, we must have $a=0$ and so from $f_{friction}=Ma$ we can conclude that $F=f_{friction}=0$ i.e. there is no force acting on the upper block. If, however, $F>0$ then we must have $F>f_{friction}$, and we need to find values of $f_{friction}$ and $a$ that satisfy the simultaneous equations $F-f_{friction} = ma \\ f_{...


3

There should be no preferential block: if the system can hold extra weight, it does not matter where it is placed; and if it cannot hold it, the system will collapse regardless. This is because given Newton's 3rd law, all stones share the same horizontal force and hence the same static friction. The assumption for that is that all the materials are the same (...


3

The blocks are a distraction. You can take an intuitive shortcut to solve the problem. Let us replace this problem with this one Either way, the total weight is $N \cdot W_{block} = W_{board}$. The upward force of friction on each end must be $W_{board}/2$. Now you are added to the board. If you are in the center, the weight on each end is $(W_{board} + W_{...


3

Just ignore the floor! If you push an object in front of you, by extending your arm, surely you and the object will have to move apart! So one or both of you will have to start moving! It makes no sense if only you move unless the other object is somehow fixed in place, since there is no critical difference between the object and you. (To see why, observe ...


3

As others said, indeed you can. This is a typical exercise in Newtonian mechanics. As @BobD said, it can be seen from Newton's third law aka action-reaction law. If your mass is $m_y$ and the mass of the object is $m_o$, both objects will experience a force of the same magnitude but in the opposite direction! The acceleration will magnitude will not be the ...


3

The table exerts a horizontal force on the block (friction). By Newton's third law, the block exerts an equal and opposite horizontal force on the table. This is the only horizontal force on table (the table is on frictionless surface), so the table moves horizontally in the direction in which you push on the block. For the sake of completeness, there are ...


3

Note that the concept of static and dynamic friction is an empirical approximation. In general, the way this is usually presented would lead one to assume that there is some discontinuous change in the forces as you go from $\vec{v} = 0$ to $\vec{v}\neq 0$. But really, the friction forces just describe the electromagnetic interactions between the ground and ...


2

If $F < f_{static}$ then both blocks move with the same acceleration $a$, so we have $F = (M+m)a \\ \displaystyle \Rightarrow a = \frac F {M+m}$ But this is the same question as you asked here. Possibly you intend to ask about the scenario $F > f_{static}(1+\frac m M)$, when the frictional force between the blocks is limited to its maximum value $f_{...


2

Consider the case when $F=f_{frictional} $. Then acceleration of the top block is 0, however the bottom block has an acceleration $a=\frac {F} {M} $. Doesn't it seem unlikely? What am I missing? What you are missing is the applied force $F$ cannot equal the friction force. When approaching problems like this it is essential to draw a free body diagram (FBD)....


2

As mentioned by silverrahul the friction force on the end block, $F_3$, is most likely to be near limiting friction. In reality the beam would bend, but if we presume that it fails only due to the limiting friction force being exceeded, it'll happen like this: Farcher found that $F_3$ is $\frac{5W}{2}$. We can get the same result by doing moments around ...


2

Here are answers to your questions in order: Slipping means $v \neq w r $. As soon as it accelerates, $v > w r $ and so it will not roll past the starting position, as $w$ will remain the same. If there's enough friction to keep it from slipping, yes. It is not conserved relative to the center of mass of the wheel if there is friction. You can use ...


2

According to Newton's 3rd law, if we apply an oblique force on a smooth surface the surface should also exert an opposite force on us. The thing is, while we are moving we don't push ourselves forward but we push the earth backwards. Think of it like you are lying on the ground and pulling a branch on the ground to move towards it. What you are actually ...


1

Static friction exists only when a body is at rest relative to the surface. Kinetic friction exists only when a body is moving relative to the surface. If a body is in motion then it will have only kinetic friction acting on it. If a body is at rest then the coefficient of static friction will be relevant. Once the force $F$ upon the body exceeds the static ...


1

There can be only kinetic friction or static friction between two surfaces in contact. The block is either moving on the surface of the plane or it is not. Once the block is moving with kinetic friction in x direction there is no static friction for a force in the y direction to exceed. So any force y will cause some lateral movement in the y direction while ...


1

Centripetal acceleration: Necessary acceleration for a body to be in circular motion. Tangential acceleration: It is because of change in magnitude of velocity. It is not a necessary acceleration for circular motion. While a car takes turn on horizontal road, it's motion now changed from translation to circular motion. We know that in circular motion ...


1

To point 1 Does slipping mean zero angular velocity or is it just the v velocity not being equal to wr? The geometric bond $v=\omega r$ tells the speed $v$ of a particle a distance $r$ from the wheel centre. Within the wheel, this equation must hold true for every particle - otherwise the wheel would break apart because particles would "skew" ...


1

Here are the answer to your doubts in order:- Slipping means there is relative motion between two surfaces. Only when $v=r\omega$, there is no relative motion between the surface of the rolling object and the surface on which it is rolling and hence no slipping. In other cases, when $v\not=r\omega$, there is relative motion between the surfaces and hence ...


1

Is the surface rough or smooth? If smooth, then friction is not present and plays no role - and I doubt you would see such skewing off from the initial direction. Such a situation would be akin to gravity-free motion through empty space; no rotation could alter the linear direction in such a scenario. (Unless we are dealing with e.g. the Coriolis effect - ...


1

Look up the impulse equation Force x time = change in momentum then imagine what would happen to you when you push the object and think how conservation of momentum can still hold.


1

First, let's comment on I know sliding is when the Vcm goes faster than the point of contact on the ground. Actually, sliding is when the point of contact on the ground has any speed (let's call this Vb from "bottom" except 0. The cases are: a. Vcm=Vb: only sliding b. Vcm>0, Vb=0: only rolling (let's call this perfect rolling) c. Vcm>0, ...


1

The definition of statics is that nothing changes and there is no speed. The Coulomb force is a phenomenon of electrostatics. Therefore, there cannot be such a thing as "a media that slows down the speed of the Coulomb force field" The only thing that might come close to your vague feeling of matter changing the properties of the Coulomb field is ...


1

My reasoning is that since at time t = 0s, the crate is still at rest, we cannot use kinetic friction since the surfaces (ground and crate) are not sliding over one another. Using this reasoning the forces acting on the crate at t = 0 must be the max static friction force and the external force. Is this correct? It depends on the exact conditions at time t=...


1

When $F=f_\mathrm{friction}$ or $F<f_\mathrm{friction}$, the friction cannot be overcome, i.e., the top block will not move (no acceleration from rest). But, the applied force $F$ will be "transferred" into the bottom block, which then experiences an acceleration $a=F/(M+m)$, since both masses act as one in this case. As an example, consider ...


1

Consider 2 blocks, A on top of B, and you apply a force on block B. The top block will start to slide relative to the bottom block when the force required to accelerate block A is greater than the maximum static friction. I.e. when $$f_{\rm A \ on \ B}>\mu N$$ where $\mu$ is the static friction coefficient between blocks A and B and $N$ is the normal ...


1

The prevailing answers seem to all suggest that friction does not stop. If that were the case, since there are no more horizontal forces, there would be some kind of horizontal acceleration due to $F_{net}=ma$, but that's not the case. The answer is that friction no longer exerts a force when the ball is rolling without slipping. Let's try to unpack that ...


Only top voted, non community-wiki answers of a minimum length are eligible