# Tag Info

8

The force is applied at the step. You apply that force downward and the step moves downward at the same time; work is force times distance (with a directional factor that is roughly 1 in this case). So the work is non-zero (and is similar to the work you would do climbing the same number of steps if the elevator was stopped.1 Now, Rahul analyzed the problem ...

4

When you are lifting an object, you are exerting a force that balances the force of gravity on the object. By $$F = m g$$ where g is the acceleration due to gravity, you see that a greater mass causes a greater gravitational force that has to be balanced by the force you apply to the object by holding it or lifting it at a constant velocity. Using the more ...

4

Work is done as long as the force is applied on the body, so in this case, the total work done would be the product of Force applied and the Displacement during the initial push only. If the object moves forever, it would do so with a constant velocity in this scenario, and consequentially its Kinetic Energy would be constant, implying that the work done ...

4

You need to look up the concept of Area Moment of Inertia, also called the second moment of area. Imagine the ruler to be made of fibers and think of a cross section to calculate the force and torque exerted by the parts of the ruler on either side of the cross section on each other through these "fibers" that run along the ruler's length, normal to the ...

3

Assuming the interpretation I suggested in the comments is correct. Consider the normal support force. It is an expression of the solidity of the surface that won't allow interpenetration. In order for penetration to not happen, there must be a force to prevent the supported object from accelerating toward the surface. Ultimately the origin of this ...

3

Why am I not accelerated by the reaction force applied by earth on me? Because the net force on your centre of mass is zero. The upward force on your feet is of the same magnitude as the downward force of gravity. Your major leg bones and spine are in compression because of the opposing forces. I know that these forces will not cancel each other ...

3

Even classically, forces arise from field being propagated at the speed of light. A physically relevant object is the energy-stress tensor, whose components represent energy density and momentum current density, so indeed momentum can be interpreted as a current that is conserved over time (as a consequence of symmetries). This point of view is also ...

3

Essentially, what you are missing is that you need to look at your situation from the point of view of the escalator steps. The escalator "sees" you "climbing" it because each step you take places you at a later escalator step. You don't keep landing on the same step. A force is applied by your foot on each escalator step. Therefore, each escalator step ...

2

Imagine a thin stick that you place on the floor and you want it stand. If you succeed to put the stick so as its weight, considered as acting on the center of mass, pass through the point $O$ of contact with the floor, the stick will stand. Otherwise it will fall. Let's see why. Let's decompose the weight of the stick into a component along the stick, and ...

2

They will experience a different force. The rope and all parts of the rope are accelerating to the right at the same rate. If you slice the rope at $P_2$, there is a tension force that is accelerating that severed section to the right. That severed section has a mass that is less than the total mass of the rope, but the acceleration is the same, so the ...

2

$F = \frac{dp}{dt}$ means that force is the rate of momentum transfer per unit time. Lets say we have mass $m_1$ moving to the right, and mass $m_2$ is on the left side of $m_1$ with zero velocity. If $m_1$ put a force to pull $m_2$, that force will create the acceleration on $m_2$ and increase its velocity, this also means the change in momentum. At the ...

2

Distance travelled can be measured as a length (meters) [$m$]. Now if we differentiate this function with respect to time ie. get the change in distance divided by the change in time we get the average speed which can be measured in distance divided by time (meters per second) [$m s^{-1}$]. If we repeat this process again we can get the acceleration which ...

2

In the framework of General Relativity, where the inertial frames are the ones in free fall, you can think that the Earth is accelerating upward, so it is not you who is pushing on Earth but it is Earth that is "running you over" because of its accelerated motion. Luckily enough, if we are standing on ground, we can avoid impulsive forces and our bodies are ...

2

In the physics definition of "work done" energy is transferred from one object (the one doing the work) to another object or system. When you push against the stationary stone you apply effort but the energy transfer is all internal to you own body - glucose being metabolized, etc. You get tired but you do no work according to the definition above.

2

When you push something and it remains at rest your muscles transfer energy through isostatic muscle contraction/respiration. This means that even though the muscles don't move they convert the glucose into respiratory energy for muscle contraction that will be dissipated eventually by heating the surroundings. The only work done is that in contracting the ...

1

The power input is roughly constant (that of a car is dictated by the total engine power while for a bicycle it depends on the user). The gear or similar tools adjusts the mechanical advantage so that a low gear will express the engine power in force rather than speed (recall that power is force times speed). On higher gears the force is traded in for speed. ...

1

Even I didn't get you but I may help you how much I can by describing your case. Your case have two bodies which are being rubbed against each other in opposite direction with constant acceleration. The definition of friction is, "The resistance which either one of the bodies offers to this motion is called the force of friction and is said to be due to ...

1

Work= force x displacement force is weight and displacement is 0 so there is no work (in physics)

1

By definition, work is $$W[\gamma] = \int_\gamma\mathbf F\cdot\mathbf v\text dt$$ i.e. force times displacement. If an object is not moving, even if subjected to a force, then there is no work done by said force. Observe that, since the object is not moving, the sum of all the forces applied to a body must be zero. For example, a body on rest on a table is ...

1

I'm assuming the circular disk is a friction-less pulley that's in equilibrium/at rest. If so N1 = N2 and the forces of tension are balanced. Once again assuming the strings are not stretchable the objects will not fall because of the tension in the string.

1

The equation governing the motion of body at right is, $m_1g-T=m_1a$ [since $F_N=ma$] ($Equation_1$) The equation governing the motion of body at left is, $T-m_2g=m_2a$ ($Equation_2$) Adding both equations we get, $a=\frac{m_1-m_2}{m_1+m_2}\times{g}$ The net acceleration produced in string at right is due to net force acting on the body at right. This ...

1

That equation expresses the fact that the forces on the block in the vertical direction add up to zero. $F_n$ and $F_{ay}$ are both forces pulling the block upwards, while the force of gravity on the block is $mg$ downwards. We don't want the block to accelerate up or down, and so the net forces in the vertical direction must cancel out.

1

When they say "Do not ignore electric force", they mean that there is both a magnetic and an electric force on the electron/positron, and you should not forget the electric force. In other words, you are asked to compute, for the $\vec v_+$, $q_+$ of the positron, the effect on the electron of its $\vec E$ and $\vec B$ field. Fortunately, 5 keV (kinetic ...

1

The force at the peg is normal to the surface of the peg - and therefore normal to the rod. In equilibrium the net torque and the net force must both be zero. You have a force (initially unknown) F at the peg - you just know the direction. There is the force of gravity on the rod - straight down, at the location of the peg. There is the reaction force of ...

1

Short answer. In your example, the rope can not be massless (otherwise its acceleration would be infinite), but if it has mass, then tension is different on each point of the rope. Therefore you have to assume there is an object connected to it. Depending on which side of the rope you assume the object is, you get tension 20N or 30N. This does not ...

1

The reaction at C is within the system. It's internal. Therefore, the free body diagram of the whole system is the whole body itself and the forces acting on the body from the outside. The reactions at A and B are external to the whole body. The force of gravity is also external to the whole body. Now, if a force were applied externally to point C, then you ...

1

If you are in orbit with this device at zero-g (free fall) then technically its not a 'pendulum' anymore. For a pendulum its assumed you have gravity as a restoring force. No gravity, no restoring force, no pendulum. It's rather a '2-body rigid dynamics' problem, but highly non-linear since the two masses will only create tension in the string when they are ...

1

The most successful attempt at a fundamental theory of the interactions in nature is the Standard Model of particle physics. In it, every interaction except for the self-interaction of Higgs is produced by it being a gauge theory, and the generalization of charge is "simply" the specification in which representation of the gauge group a given field ...

1

The direction of kinetic friction is in the opposite direction of the velocity of the box and independent from acceleration thus whether the $\mathbf a$ is positive, negative or zero does not affect the direction of the friction vector

1

Natural frequency depends on the physical properties of a system. Some of the classical examples are masses attached to springs and pendula. For the former class, the basic model is based on Hooke's law, which translates into the differential equation (1D) $$m\ddot x + kx = 0,\qquad m,k > 0$$ where any sort of dissipative effect has been neglected. ...

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