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For a pulley that has mass and moment of inertia, there must be a net torque on the pulley for the pulley to demonstrate an angular acceleration. Assuming that mass "M" is greater than mass "m", a net torque necessarily requires that the counterclockwise torque from mass "M", given by the equation Torque1 = T1(R), is larger than the clockwise torque from ...

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Yes. Tension can vary if external forces are acting between the ends of the string - such as gravity (if the string has mass) and friction where the string makes contact with other objects (such as the pulley). For example, suppose you attach one end A of a uniform massless string to a support and the other end C to a vertically hanging mass M. This ...

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It seems to me that the key to this trick is to build up enough elastic energy in the rope - which requires you to "build tension" by riding the edge hard, as explained in this video. If you do the trick too close to point A, there is limited lateral motion needed to build tension - but as you take off, the force you are looking for will disappear as the ...

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We know $$f_n=\frac1{2\pi}\sqrt{\frac km}$$ We then ask ourselves what is k and m. For this case, mass is $980kg$. Is k stiffness of 1 spring or 4 spring? The 980kg mass is not sitting on 1 spring. So it should be 4 spring. The tricky part is we don't use 4k. Instead we use k for stiffness equivalent to 4 springs. With 80kg, we get 1.2cm deflection. So it ...

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You need the effective spring constant of the four springs in parallel and then use the whole mass of the car to consider the motion of the car or use the spring constant of one spring and use a quarter of the mass of the car.

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The answer has pretty much been given in the comments, but I think a nice pictorial representation might help. The mathematical form of a standing wave is $$y(x) = \sin \left(\frac{2 \pi}{ \lambda} x \right)$$ Here $y(x)$ is the displacement of the string at point $x$. If we plot the waves for the four wavelengths we obtain the following picture I will ...

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The tension in a string is uniform Not always. The tension of the string is uniform in some cases: If String is mass-less and its particles don't move with respect to each other (i.e. string is inextensible or if it is extensible, it reach to its final tension). If String is mass-less and there is no friction between string and pulley. For string with ...

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The basic problem is that headphones are heavy and asymmetric, whereas there's nothing attached to the shoelace. What this means is that the headphone - cord system can get stuck in stationary, locally but not globally minimum energy configurations: the twisting of a cord raises the energy of the cord, but the torsion resulting from the twist is not enough ...

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I think the problem consists three assumptions: The rope is mass-less. The rope is inelastic. There is no friction between rope and pulleys. So, you can determine the tension of the rope by considering to its free body diagram: If we isolate an arbitrary infinitesimal element of the rope, then the forces those acting along the element length must ...

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I think you are confusing static and dynamic or are trying to separate the two in a way that isn't helpful and leads to 'over-thinking' of a simple problem. Let's look at the force diagram again: In the point $P$ a downward (but as you stated not necessarily vertical) force $\vec{F}$ acts on the end of the string. We decompose it into two components: $\... 1 This might further clarify (or confuse) things: The force that an object exerts on the string is equal in magnitude to the force that the string exerts on the object. The force that an object exerts on the string (or similarly the force that the string exerts on the object) is not necessarily equal in magnitude to the component of a force acting on the ... 1 Your statement is quite a complex one, involving several clauses. It is not clear which clause is the focus of your question. You state that the string is vertical and the applied force is downward, so the force$F$has no component perpendicular to the string, only in the direction of the string. You do not explain what you mean by the "constraint ... 2 You would be right, if the string is like a real string in that it's flexible (that is, can neither exert transverse force nor bear transverse loads). I have to point out here that you haven't described a system (classically, you couldn't draw a static free body diagram for the string)--to be static, the entire force$\vec{F}$would have to be parallel to$\...

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