# Tag Info

56

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 ...

17

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 ...

13

This is the boring answer that your physics teacher would give you :-) The diagram above shows an idealised rope with all it's weight concentrated at the centre of the rope. If the rope mass is $m$ then the force downwards is just $mg$, where $g$ is the acceleration due to gravity. Suppose you're pulling on the rope with force $F$, and the angle of sag ...

13

I believe the following diagram tells you everything you need to know: when the center of mass is above the support point, then the lower point will carry more of the weight since it is acting closer to the center of mass ($x_1 < x_2$) and torque balance requires that $F_1 x_1 = F_2 x_2$. Conversely, if the center of mass is below the support, the higher ...

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 ...

8

There are three parts to this: In mechanical equilibrium, things go to their lowest energy state. A straight line is the shortest distance between two points. Whenever you've minimized something, it means that small deviations don't change its value (to first order). Let's start with the third point, which is mathematical, and then look at the physics of ...

7

The horizantal beam on such scales is intentionally placed below the rotational axis. As long as the weights are in equilibrium the torque is equal on both sides. But as soon as the position changes e.g. tipping the left scale down, the torques differ because only the tangential part of the gravitational force vector in relation to the rotational axis ...

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 ...

7

It took a while, but I found a good reference on this problem: A study of the generalized catenary problem (PDF), an undergraduate mathematical physics thesis by Hai Xuan Nguyen. It explores the equations governing the equilibrium shapes of chains and arches in radial potentials $V(r)$. I've extracted and adapted the relevant parts for this answer, and ...

6

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 ...

6

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. ...

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

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 ...

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 ...

6

Consider this for the upper ball: The angle of the inclined plane is simply derived from geometrical considerations. Looks pretty easy to solve, right? Once you know the normal force of the inclined slope, reflect and apply it to the bottom ball to complete the problem.

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

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

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

If it would only be the weights exerting torque, the balance would be in equilibrium at all angles. What makes the balance go back to the horizontal position is the fact, that the center of mass is below the beam. consider this picture The needle exerts a torque too, so you have more torque on the side, where the plate is higher. You can have more subtle ...

4

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 ...

4

You can't pull a piece of string perfectly straight - you just can't see the sag. (excluding the limit where variations in the thickness of the string are greater than the sag) Two simple ways of looking at it: Mathematically - As you pull tighter the sag gets less, but it's a function of 1/force, so to get zero sag you need infinite force. ...

4

Here is an idea, borrowed from linear elastic theory. Solve all the forces in terms of an unknown force (I chose $f_{10}$) and construct a long vector $f$ $$\boldsymbol f = \begin{bmatrix} f_1 & f_2 & \dots & f_9 \end{bmatrix}^\top$$ where each component is a function of $P$, $\beta$ and $f_{10}$. Now assemble something resembling the total ...

4

There is no "correct" point. Using any point will give you the same answer. If you didn't get the same answer using both points then you've done something wrong. Hint: you can treat the gravitational force as though it were acting at the shape's center of mass.

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 /...

4

Here's a motivation for where the inertia tensor $I=(I_{ij})$ (and by extension moments of inertia) comes from. It's a quantity that's analogous to mass for rotational motion in the sense that the kinetic energy of a rotating object is essentially proportional to the inertia tensor times the square of the body's angular velocity. More precisely \begin{...

4

Although Floris made a clear picture involving only vertical forces, this picture is mostly useful when carrying washing machines or large chairs, where the 'height' of the object is more pronounced. However, you will find that, even when the object is mostly flat, the bottom person will carry most of the weight. The key here is the direction of the forces ...

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