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47

This is a statics problem. Assume the cable is static, perfectly straight and horizontal. Pick any point on the cable and the sum of the forces on that point must equal zero. There is a force, due to gravity, "downward". So, there must be an equal, opposing force "upward". This upward force must come from the tension in the cable. But, if the cable is ...


17

To make it fall you need a torque. This torque is provided by the weight force acting on the center of mass of the object and by the offset between the center of mass and the edge of the object. Imagine your domino standing upright then tilt it. You are moving the center of mass. When the center of mass (blue) is on the right of the edge (red) then you have ...


12

Imagine a heavy chord raised off the ground between two blocks. Rather than consider all of the mass pieces of the rope, and the forces on them, we can simplify the problem a little bit by considering a slightly different one. The chord can be represented by a heavy ball (in the middle of the chord) connected by two massless strings to the blocks. From ...


11

Newton's first law of motion for a point particle states that a particle at rest will stay at rest and a particle in motion will stay in motion unless acted on by an unbalanced force. In other words, if the net force on the particle is zero, then the velocity of the particle will stay constant. Newton's first law of motion for a system of particles states ...


7

Yes, it's called the normal force. It comes from the rigidity of the stuff separating the object from the center of gravitational attraction, i.e. the rigidity of the rocks, dirt, floor, table, etc. If you'd like, you could think of this stuff as behaving like a spring with a huge spring constant. Any first-year physics textbook will cover this; there's a ...


6

This should be possible to solve in the same way we do the ordinary catenary problem, by variational calculus. Suppose the angular separation between the endpoints is $\Delta$, where we could define $\Delta = \frac{D}{R}$ if I understand the problem correctly. Let the shape of the rope be given by a function $r(\phi)$, and write the potential energy of the ...


6

Kind of. The negative sign indicates the direction of the force exerted by the spring on the mass. If you pull the mass to the right, the force from the spring is to the left. Since they go opposite directions, there is a minus sign. The problem states an external force exerted on the mass displaces it, presumably to a new equilibrium. The spring ...


6

Since this is a homework-type problem, here are some Hints for the force The electrostatic force $d\vec F$ on a small segment $dl$ of the rod given the field $\vec E$ of the other rod is $$ d\vec F = \lambda\, dl \,\vec E $$ Determine the field of one rod, and use the above expression to integrate the force it exerts on the other rod. This is a 2D ...


6

As you have noticed yourself, your system is simply underdetermined. In order to find a unique solution you need to add some extra constraints in addition to Newton's equations. Imagine a table with more than four legs: the more legs you add, the more unknown forces you have. But the number of equations does not change. If we instead remove a leg we find a ...


5

Yes, it's possible. A static setup like this will work as long as any small motion of the parts would increase the potential energy. In this case, it looks like there is only one possible motion - rotation of the entire ruler-hanger-hammer piece about the axis where the ruler touches the table. If the ruler were to rotate down a little bit, the entire ...


5

A push up is a form of lever. The athlete must exert roughly half her body weight (under some assumptions I'll clarify at the end of the post.) We can solve this problem using the principle of virtual work. Assume the athlete raises her body through a small angle $\textrm{d}\theta$. Then her center of mass rises by $l \cos\theta \ \textrm{d}\theta$, with ...


5

It is hard to guess without seeing Gorillapod in use, but my guess would be the following: Center of mass could be understood as an average position of the mass of the object. In order for an object to be in stable equilibrium, its center of muss must be vertically above the area, which is enclosed by contact points of tripod's legs with the ground. If ...


5

Some engineering texts use "moment" and "couple" to talk about forces that tend to rotate an assembly (what physicist mean when they say "torque", but the engineers sometimes have a slightly different meaning for that word). A roughly translation guide is... A "couple" is a pair of opposite forces whose points of action are not co-linear. A couple is ...


5

The simple answer is that you can't fully solve this problem--because as you note it is under-constrained--under the assumptions that are made when you first start doing statics (that objects are completely rigid). The introduction of finite strains bring in additional relationships.


5

These are some of the Newtonian couples. The weight pulls down on the rope, and the rope pulls up on the weight(tension). The rope pulls down on the pulley(tension), and the pulley pulls up on the rope. The pulley pulls right on the rope , and the rope pulls left on the pulley(tension). the rope pulls right on the frame (tension), and the frame pulls left on ...


4

It's a 20-lbs staff. 20-lbs net force are required to hold it up, regardless of its orientation. If you just apply this force to one end of the staff, though, there would be a net torque. Instead, you need to use your hands to apply two forces to the staff. One force, exerted by the near hand on the very end of the staff, should be down. The other, ...


4

As joshphysics' answer showed, the force would indeed be infinite in the case of uniform lineic distributions, but the torque does not need to be. Using the same conventions joshphysics did, let's compute the elementary torque $d\vec{T}$ experienced by a piece $dr_2$ of rod 2 at position $\vec{r_2}$ from the joining point and integrate it over rod 2. It can ...


4

This looks like a simple linear blending problem. It is two-dimensional, but each dimension can be considered independently. The more to the right the weight is, the larger the fraction of it carried by F2 and F3. Basically, the fraction of the weight carried by F2 and F3 is X/W. Put more mathematically:    (F2 + F3) / (F1 + F2 + F3 + F4) = X ...


3

The book is correct - how many significant figures are you given the data to ? I would probably use the middle of the supports (ie 16m) but that doesn't matter for working out the vertical forces. This question is also nothing to do with momentum, unless there is a part 2 where you work out the sideways force when the truck moves.


3

I know this is an old question, but for the benefit of people visiting here wondering what the answer was, here it goes: A droplet can stay at rest on an inclined plate because of small heterogeneities on the surface. This can either be a small roughness (of the order of nano/micrometers) or `dirty' spots where the surface chemistry is locally different. ...


3

$\newcommand{\t}[1]{\frac{T_#1}{\sin\theta_#1}}$ The correct way is with vectors. The shortcut way is to use Lami's theorem. Basically, if the tension in each string is $T_1$, and the angle opposite each string is $\theta_1$, then $$\t 1=\t 2=\t 3$$ Referring to this diagram, $\frac{A}{\sin \alpha}=\frac{B}{\sin \beta}=\frac{C}{\sin \gamma}$, where ...


3

Let $y(x)$ be the curve describing the shape of the cable. Let $T(x)$ be the tension in the cable. Consider a small segment extending from $x$ to $x+dx$. The horizontal component of the tension at the two ends of this segment must cancel out, so $T_x$ must be constant. If $\theta$ is the angle the cable makes with the vertical, so that $y'=\tan\theta$, ...


3

Start with differential form of Poisson's ratio: $$\frac{\text{d} x}{x}=- \nu \frac{\text{d} l}{l}$$ $$\int_{x_0}^{x_0+\Delta x} \frac{\text{d} x}{x}=- \nu \int_{l_0}^{l_0+\Delta l} \frac{\text{d} l}{l}$$ $$\ln \frac{x_0+\Delta x}{x_0}=- \nu \ln \frac{l_0+\Delta l}{l_0}$$ $$1 + \frac{\Delta x}{x_0}=\left(1+\dfrac{\Delta l}{l_0}\right)^{-\nu}$$


3

This is blatantly a homework question, so we're only allowed to discuss methods, and not give you the answer. With any problem like this the very first thing to do is draw a diagram. From the limited information in your question I think the situation loks like this: You know the force $F$ that you're using to lift the end of the plate, and you want to ...


3

Continuing from Pygmalion's answer a graphical explaination can be as below (Warning! - Representation may be a bit wierd and out of proportion) Initially without the camera this is the case After camera with long lenses is placed the COM (centre of mass) shifts upwards and outwards as below This might be the "Centre of Mass" issue you are talking ...


3

Suppose you have a balance that looks like this: where $m_1$ and $m_2$ are the weights you're comparing and $M$ is some large weight fixed to the balance. For simplicity let's $m_1$ is zero, so one one end of your scales you have some weight $m_2$ and there is nothing else on the other end. You are quite correct that with simple lever scales the lever ...


3

As others have pointed out in the comments, it is not really trivial to apply any model to reality, especially since we don’t know much about reality. However, we can make a few educated guesses and estimations and see how well everything fits together. First, assuming perfect weight distribution, we see that five screws would probably be sufficient to ...


3

It's pretty simple: the forces at the anchor points would be infinite because of the 90° angle ;-) An example: Imagine two pillars with the same height. If you attach a rope on both of them and try to tighten it, you will slowly increase the pulling force at the top of the pillars while increasing the angle between rope and pillar. To fully straighten the ...


3

The pieces that support the most weight have higher friction and are more difficult to remove. The easier it is to remove a piece the less important it is structurally. Each block needs to support the weight of all the blocks above it, and it has to have at least 3 contact points spread apart like a three legged chair. With two contacts points it will create ...


3

The motion will depend on the forces that the torque arises from in the first place. Newton's second law must apply to the system's centre of mass and is of course reckoned with the nett force. So if this nett force is nought, the centre of mass either cannot shift or moves uniformly (if so we shall assume the sphere's centre of mass is stationary ...



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