My biceps muscles are attached to my arms bone/bones through a tendon/tendons, if my biceps pulls my arms bone/bones with a force of 100N upwards then my arms bone/bones pushes back downwards with a force of -100N,therefore fnet should be 0 and there should be no movement, however in reality our arm moves upwards, why?
4352
Given Newton's third law, why is there motion at all? Should not all forces even themselves out, so nothing moves at all? When I push a table using my finger, the table applies the same force onto my finger like my finger does on the table just with an opposing direction, nothing happens except that I feel the opposing force. But why can I push a box on a table by applying force ($F=ma$) on one side, obviously outbalancing the force the box has on my finger and at the same time outbalancing the friction the box has on the table? I obviously have the greater mass and acceleration as for example the matchbox on the table and thusly I can move it, but shouldn't the third law prevent that from even happening? Shouldn't the matchbox just accommodate to said force and applying the same force to me in opposing direction?
51918
Am I correct to say that when a human lifts a dumbbell from, say, 2 feet off the ground to 6 feet off the ground, he will have increased the potential energy of the weight and thus will have burned calories to supply that energy increase?   Will two different people, Arnold Schwarzenegger in his prime and a scrawny kid for example, expend the same amount of calories moving the dumbbell from 2 feet straight up to 6 feet? Even though it's much easier for Arnold to lift it up, is he still burning the same number of calories as the scrawny kid?  If one lifts the weight up very slowly, isn't that going to expend more calories? When you prolong your exercise, but still do the same number of repetitions, you sweat a lot more and feel more tired, which seems to say that more calories are burned. But there's no time variable in the formula for potential energy (U = mgh). What other energy equation must we use then?
68225
This is really bothering me for a long time, because the math is easy to do, but it's still unintuitive for me.  I understand the "law of the lever" and I can do the math and use the torques, or conservation of energy. or whatever... And I can see that a lever can amplify a force you apply to it  if you apply a force on the longer side of the beam.  If I were to look at the molecular lever and see what actually happens when I push on the lever, and I give acceleration to the molecules, how does it actually happen that more force is transmitted to the other side?  Thank you all  p.s I'm looking just for an explanation in terms of forces and acceleration, it's clear to me how to do this in terms of energy or torques
19257
I have learned that $$\mathbf{F} = m \mathbf{a}$$ where both the force $\mathbf{F}$ and acceleration $\mathbf{a}$ are vectors. This makes sense, since both force and acceleration have a direction.  On the other hand, the kinetic energy $$K = \frac12 mv^2$$ looks completely different. It doesn't seem to depend on the direction. How are these two concepts related?
6838
I am aware of an intuitive explanation  for $\operatorname{curl} \operatorname{grad} F = 0$ (a block placed on a mountainous frictionless surface will slide to lower ground without spinning), and was wondering if there were a similar explanation for $\operatorname{div} \operatorname{curl} F = 0$.
112540
Here's an image from another question:    I get what the pull down resistor is there to do, but I don't really understand exactly how it works.  When there is no voltage applied through the base wire, why does any induced current then go through R2, but when a voltage is applied, it goes to the transistor? My understanding is that voltage always follows the path of least resistance, so if applying say, 5v turns the transistor on, that means that the transistor has a lower resistance than R2, so I would also assume any induced current would also flow through the resistor - so why does it then go through R2?
33751
When solving for the force component ($F$) in the direction of the string, the following triangle is drawn.    (Adapted from )  Giving $F=mg * cos θ$  Why is it drawn that way, and not like this?    Which would instead give:  $F=mg / cos θ$  $ cos θ=mg/F$  $F=mg / cos θ$
12911
If the friction on an object is 200 N and you apply 200 N, it seems, the forces should immediately cancel each other out since the friction and your applied force go in opposite directions, leaving you with an object that is motionless. Yet, from what I've read, that's not the case. I feel like I'm misunderstanding something. Could it be I'm wrong about needing to apply 200 N initially? Might the amount of force I actually need be greater, but instantaneously decrease to 200 N?
73298
This is the question:      A rope brings a 50 kg object downward at an acceleration of 0.75. What is the tension force of the rope?    I thought that, since the tension and the weight are both "going" downward, the net force would be Weight + Tension = Mass x -Acceleration. The acceleration is negative because it is going downward. But if I do it that way I get the wrong answer. The answer is only correct if net force is Weight - Tension, and if acceleration is positive, neither of which makes sense to me. Can someone explain this?   It could be a silly arithmetic error that I haven't noticed.
33025
A $ 2.0 kg $ box of cucumber extract is being pulled across a frictionless table by a rope at an angle $ \theta=60° $ (from positive direction of $ x $ axis, we have taken horizontal surface of table as $ x $ axis) The tension in the rope is $ 12N $ and causes the box to slide across the table to the right with an acceleration of $ 3.0 m/s² $ But the direction of net force is along the rope. The direction of acceleration isn't. Why is that?
101333
Could someone give a (detailed) explanation (perhaps with formulas) of how the load (torque) affects the armature current of a DC motor (separately excited or shunt or series)?  As the load torque increases the speed of the motor decreases, and assuming the terminal voltage stays constant, the EMF will become lower and thus the armature current will increase.  But could someone give me the formula that shows the relation between load torque and armature current?  Is it something like:   \$T_\text{developed} = T_\text{shaft} + T_\text{friction windage} + T_\text{load}\$ ,   where \$T_\text{developed} = K \, \text{flux}_\text{pole} \,I_\text{armature}\$ ?  But then you have  \$P_d = P_s + P_{f_w} + P_l   \$  where \$P_d = K \,\text{flux}_\text{pole}\, I_\text{armature}  \,\omega_m\$  In the last formula you can't see that as the increase due to \$P_\text{load }\$ the armature current increases because maybe someone can say the speed \$\omega_m\$ increases.  And how come at no-load the armature current and thus the torque developed is zero? Is it really zero, or do I have to assume there is some current flowing?  Could also explain me the principle of torque developed. Because at no load, the shaft torque is equal to torque developed, right?
73118
I tried doing  just for fun. It is reproduced as follows:          (I'm not allowed to post the entire problem here since I'm a new user.)  Here are my equations:  $$T - mg = ma$$ (equation 1)  $$Mg - T = Ma$$  (equation 2)  (Upper case "M"  used for infinite series of masses on the right arm of the first pulley. Lower case "m" for the small, single mass on the left arm.)  algebra is as follows:  $$Mg - (ma + mg) = Ma$$ $$g (M - m) = a (M + m)$$ $$a = g (M - m) / (M + m)$$   Here is my reasoning to get final answer:   The limit of $a$ for large $M$ goes to $g$. Therefore, acceleration of the top mass is $g$. However, the correct answer is $g/2$. Why does my reasoning get the wrong answer? Any ideas?    (Please excuse the formatting, I'm a newbie here).
102620
suppose i jump from ground then work done by normal is zero as point of application  displacement is zero(work is doneby internal forces) but during the pushing part  COM has upwards displacement then shouldn't work be done if apply force on COM  ??
95828
I know this has been asked around, my question is a bit more specialized towards my personal problem.  If I have an object the magnitude of the weight is usually equal to the normal force, right?   I could be mistaken but from what I have read the normal force is the force the object is feeling opposite to the $(mass*gravity)$. If that is the case, and I put my hand on the object does my Force get added to the object's weight and thus the normal force of that object increases? Therefore if I keep adding hands to the object will the normal force keep compensating?   Also is the normal force the negative charged electrons repelling each other (electromagnetism)?
13179
I'm having some trouble understanding the concept of negative work. For example, my book says that if I lower a box to the ground, the box does positive work on my hands and my hands do negative work on the box. So, if work occurs when a force causes displacement, how does negative work happen? Are my hands displacing anything?
49742
I'm a high school student and I'm studying Newton's second Law. While my teacher is calculating a net force of an object he always treat all vectors as positive numbers. I think what he meant is the magnitude of the vector which is always positive, but the problem is that he didn't use the magnitude symbol, for example like $||Fg||$. He said the acceleration due to gravity , g, is positive, but shouldn't it be negative since it is pointing down. This also cause a problem: every time when I calculate the net force I can't add all forces together, instead I need to choose to use minus sign when the vector is pointing down. So in general which way is correct?
5257
As a layman, i can calculate approx "displacement" just by observing the moving object. And accurately by using a simple "scale". Similarly, again, I can calculate angle from origin by using displacement in $x$ and $y$ dimensions. Similarly I can use a stopwatch and scale to understand velocity.  But when I read about , first of all it confuses with the english word "force", we use in real life. To some extent I am sure, it has nothing to do with that. So exactly what is it in the real sense. How can a layman, see force as, just like he can see displacement or angle? Or force is just a quantity defined by physicists to simplify the combinations of $ma$, they might be facing every time. And thus came up with term "force" ( which is similar in spelling to english word in oxford dictionary). And lastly, why not force has just been called as something proportional to mass and it's displacement or velocity. Why something at the level of change of velocity has been used to define it.