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## New answers tagged newtonian-mechanics

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First remember that Air resistance is proportional to $v_{net}^2$. And $v_{net}^{\text{freefall(1)}}<v_{net}^{\text{throw at angle$\theta$to normal(2)}}$ So air-resistive force acting in free fall proportional to $(v_{net}^{(1)})^2$ And air-resistive force acting in the other case proportional to $(v_{net}^{(2)})^2\cos(\phi)$ ($\phi$ is the angle ...

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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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Relative motion between block and belt will stop when speeds of block and belt become same. So, after finding the acceleration of the block due to friction, you should find the time taken for it to reach the speed of the belt. Note that, speed of the belt, is constant.

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Friction force always opposes relative motion between two surfaces. In most cases (like yours), one of those surfaces is fixed. So, you should recognize that how does the other surface tend to move. For this purpose, you should consider that what force or torque want to move the body. Then, you can determine the correct direction of friction. While ...

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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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Remember that static friction is a constraint force: it enforces the rule "no motion between these surfaces" as long as the force needed to do so does not exceed the maximum. The force of static friction will have the value needed to prevent relative motion unless that value exceeds the maximum. So there are two possibilities to what happens here. ...

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When $F=2N$, the a friction of $2N$ would also act in the opposite direction on $A$ This is wrong. Friction force isn't equal to force $F$. It is equal to $F_f=m_Ba_B$ Free body diagrams of blocks are as below: When $F=2\;\mathrm N$ then we have: $$F-F_f=m_Aa\Longrightarrow2-F_f=2a$$ $$F_f=m_Ba\Longrightarrow F_f=4a$$ Hence $$a=\frac 13\;\mathrm{m/s^2}... 2 Your reasoning is essentially correct, apart from the last paragraph. To conclude, note that Newton's equation:$$\ddot {\mathbf r}(t) = \mathbf f(\mathbf r (t)),$$with initial condition \mathbf x (0)=(x_0,0,0), \dot {\mathbf x} (0)=(0,0,0) can be solved by puttin y(t)=z(t)\equiv 0, thus reducing to a one dimensional problem:$$\ddot x (t)=f(x(t)),$$... 1 We know$$f_n=\frac1{2\pi}\sqrt{\frac km}$$We then ask ourselves what is k and m. For this case, mass is 980kg. Is k stiffness of 1 spring or 4 spring? The 980kg mass is not sitting on 1 spring. So it should be 4 spring. The tricky part is we don't use 4k. Instead we use k for stiffness equivalent to 4 springs. With 80kg, we get 1.2cm deflection. So it ... 2 Mass removal due to wear relates to sliding distance. This gives to wear rate relates to sliding speed. However, wear rate is not only only a function of sliding speed. Surface hardness also plays a role. AL-SI, according to this paper, will be hardened in the beginning session. With the surface hardness increased, the wear rate decreases. When it can no ... 0 I shouldn't give a direct answer since this is an exercise question. Since the System (Man+seat) is rotating on a parallel to the ground plane. That means that there is no motion on the vertical axis-y. That means that the total force in this axis is zero.$$ \sum \vec F_y = \vec 0$$You have to find what are the forces on this axis. It can easily be done ... 0 What your book meant was, "When applying the equations of motion, it is important that the acceleration of a particle be measured by an observer traveling in a reference frame that is either fixed (in space) or translates with a constant velocity'. The n-t coordinate system is fixed in space, and is not traveling along with the particle. The reference ... 0 I assume that air resistance force is parallel to the velocity vector but in opposite direction. We have:$$a_x=10\cos\thetaa_y=10\sin\theta-g\;\Longrightarrow\;a_y+g=10\sin\theta\tan\theta=\large{\frac{v_y}{v_x}}$$Then,$$\large{\frac{v_y}{v_x}}=\large{\frac{a_y+g}{a_x}}\;\Longrightarrow\;\large{\frac{a_y+g}{v_y}}=\large{\frac{a_x}{v_x}}=C\;\...

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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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If we are at the equator on the surface of an atmosphere-free, non-rotating, perfectly spherical planet (no mountains or trees, etc) then the optimal angle should be zero (launch towards the horizon, tangent to the planet's surface). This assumes the initial velocity is the orbital speed at the surface of the planet, which is approximately the escape ...

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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 first thing first -The concept of friction Friction come in play when there is any tendency of relative motion between 2 surfaces and it is in opposite direction of relative motion (Note- I used word relative motion not simply motion). In simple word friction try to reduce relative motion. in 2 block problem and this problem the difference is T (tension)....

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When you jump from a height, you gather momentum. Absorbing this momentum at landing reduces the size of the maximum force, and thus the "pain". Let us assume that the distance over which a person can absorb the momentum of the fall is proportional to their height (proportional to the length of their legs). In that case, the taller person can absorb the ...

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I question the physical possibility of the question as posed. If A is accelerating to the left, as I believe it would, then the pulley is accelerating to the left as well. Then B, hanging from the pulley, must accelerate to the left (as well as down, of course). Otherwise it would be "left behind" by the block A. The only possible source of a horizontal ...

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It is not true that the same force has to create the same change in kinetic energy. For instance, if two equal forces of opposite directions are applied on a body, the body does not change its energy. Thus each force makes zero work, or zero change in kinetic energy. You could tell that both forces create kinetic energies in different directions and that ...

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Energy is force times distance. You haven't specified a distance, so you can't say how much energy that force delivers. If you allow the force to rotate the sphere, the application point will accelerate faster than it would if the sphere does not rotate. So a constant force will deliver energy to the sphere faster than it would in the non-rotating case. ...

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I think you'd better take a clear force analysis first. This is not a hard problem, though it may take a few tricks. There's some hints for you: Take the pulley and A as a whole; Investigate carefully on the direction of every elastic force; A and B move simultaneously. The answer consists only of m(A), m(B) and g.

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I am assuming that A is a free body on a frictionless plane and that the rope is fixed to the wall to the left. The key concept to understand here is that the tension in the rope as a consequence of the weight of B must be equal at all points. You already seem to have understood that concept. However, you state that the two forces are in the same direction. ...

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The arrow ideally will fly out like a drag race car with the parachute deployed! A well designed arrow should have these properties. 1- Sharp and proportionally heavy point to accept a large momentum and deliver it as kinetic energy E= mv^2/2 2- long and balanced stem to accommodate a big arch and maintain separation between the tip and the fletching. 3- an ...

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We must all keep in mind that for average atmospheric pressure, and assuming Zhang could pull a hard vacuum with his abdomen (which is probably not feasible, but serves to provide us with a bound), the maximum (negative) pressure he could achieve is only about 14.7 psia. Given 36 tonnes, you can back calculate what the diameter of the bowl would have to be. ...

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In the movies, arrows shot into the air rotate so that during the descent, the arrow head hits ground first. What is the source of this angular momentum? An arrow shot on the Moon would not do that. Air and the geometry of the arrow are key. An arrow flying through the air is subject to two forces, gravity and aerodynamic drag. Gravitation will not make an ...

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I draw one for you,hope it helps!

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What should be the minimum value of $v_0$ in order to hit the monkey while it's in air? Minimum value for $v_0$ is when arrow hits to the monkey just before it (monkey) reaches to the ground. Or, minimum value for $v_0$ is when arrow's range is equal to horizontal distance between hunter and monkey. So you need to find the time of monkey's fall. Or, you ...

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The exact mechanism you describe for how suction cups work is how the rice bowl work. Instead of the bowl being flexible, though, it's his body (skin and muscles) that are providing the change in volume necessary for the suction. So, instead of the suction cup creating the volume change, it's the surface the suction cup (the bowl) is attached to, Mr. Zhang'...

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The torque supplied by $F_m$ results in the torque due to $F_v$, so these torques are equal : $\vec {JV} \times \vec F_v = \vec {JM} \times \vec F_m$. Evaluation : (a) either $\vec A \times \vec B = (AB \sin\theta) \hat k$ where $A$, $B$ are magnitudes and $\theta$ is the angle between (b) or $(A_x \hat i+A_y\hat j) \times (B_x\hat i+B_y\hat j) = (A_xB_y ... 0 You can easily find direction of friction force by drawing free body diagram for the cylinder. FBD of the cylinder is shown below (Note that I have drawn diagrams for counter clockwise rotation) As you see in figure above, friction force$f$opposes to rotation of the cylinder. Equations of cylinder motion are as below $$f=ma_G\;\tag 1$$ $$N=mg+F\;\tag 2... 0 If the cylinder is not accelerating and the rolling no slip condition v_{cm} = R \omega is satisfied then the frictional force between the cylinder and the ground is zero. The horizontal force on the cylinder is zero as it is not accelerating in that direction. The weight of the cylinder is equal an opposite to the normal reaction pn the cylinder due to ... 44 Air. Conservation of angular momentum does infact dictate that whatever rotation it starts with it should end with, provided nothing else acts on it. Air allows its forward momentum to act on it. Consider a weather vane, a wind sock, or a flag. They rotate when not facing into the wind because one side presents more wind resistance than the other. Once ... 8 As pointed out by dmckee in his comment, anyone (including myself) that has practised bow and arrow knows the arrow by weight and fletch inspection. As my English corrector points out, fletch doesn't seem to be a very commom word: it means that feather at the end of the dart/arrow. Everything resumes to how good is the fletching. When shot the arrow wants, ... 49 The same reason objects which are heavier on one side tend to fall with the heavy side down: the tip of the arrow is denser than the rest of the arrow. The center of gravity is offset from its geometrical center, so the air drag, which is based on the object's geometry, causes a torque together with gravity as seen in this very professional picture of a body ... 2 Newton's second law, force f is$$f=m\frac{d^2 x}{d t^2}$$x is position vector of the particle.$$f=-\frac{d v}{dx}$$v is the potential energy.$$m\frac{d^2 x}{d t^2}=-\frac{d v}{dx}$$Multiply both sides with \dot x$$\frac{m}{2} \frac{d\dot x^2}{dt}=-\frac{dv}{dt} \frac{d}{dt}(\frac{1}{2}m\dot x^2+v)=0$$i.e.,$$\frac{dE}{dt}=0$$Energy is ... 2 The flaw is your assumption that In this [accelerating] frame we don't see any force so the first law of dynamics is respected. In the accelerating reference frame you do see evidence of a force, even though you don't see the effect you are expecting (acceleration of the object). Like an observer standing on the surface of the Earth, acted on by the ... 0 First of all, I shall assume that the system is kept on a horizontal plane so as to simplify the calculations. Secondly, I shall assume that the line joining the mass m and M is perpendicular to the direction of motion of mass 3m. Thirdly, I shall assume that the spheres to be point objects (in other words, you can consider it to be a perfect head-on ... 2 No this is not a physically valid equation. It is a mathematical description of the pendulum in which the variables do not have units. The corresponding physical equation would be:$$I \frac{d^2\theta}{dt^2} + b \frac{d\theta}{dt} + c \sin(\theta) = T \sin(2\pi ft),$$where$$[I]=kgm^2, [b] = Nms, [c] = [T] = Nm$$For small angles$$ sin(\theta) = \... 1 Due to the fact that the body in equilibrium, all forces must cancel each other. Which forces must cancel each other when a body is in equilibrium? When a body is in equilibrium, resultant of forces acting on it must be zero. In current question, we have three bodies those are in equilibrium (man, pen and the earth). So, net force exerted on each of them ... 5 Newton's third law tells us that the momentum imparted on one body is equal and opposite to the momentum imparted on another if they interact. We then have $$\Delta \vec p_1~=~-\Delta\vec p_2.$$ The change in momentum is$\Delta \vec p_i~=~m\vec a_i\Delta t$,$i~=~1,~2$. The change in momentum is with Newton's second law due to a force so that $$\vec F_1~... 0 You need to consider the moments of the forces involved. Moment is force times perpendicular distance from the axis. In order to turn the dumper bucket, the clockwise moment of the force from the hydraulic cylinder needs to be at least as much as the anticlockwise moment of the bucket's weight (when loaded). The weight of the bucket acts through its centre ... 7 The zig-zag strategy seems trivially obvious - but it might not be the better strategy in a particular situation. I suggest rather than asking under what conditions this strategy is preferable, you ask under what conditions the counter-intuitive straight-line strategy is preferable. The advantage of zig-zagging is that it presents a smaller "collision ... 2 if the body is rolling on a plane, then its degree of freedom is 2: one for rotation about the body's axis and one for translation of its center of gravity in forward and backward direction. if there is no slipping, the translation can be calculated from the rotation and the radius of the block. Thus the degree of freedom is degenerated to 1. 7 You need a better statement of Newton's Law. The one you are using is meaningless, because the word action is not defined. (In today's language of physics the word action is used in an entirely different context, sense, and meaning.) It's based on what Newton wrote, but is only half of what he wrote. Wikipedia gives us the whole thing To every ... 46 The way we are all taught Newton's Laws (by reciting them like mantras as children) is unfortunate because the traditional wording is misleading in many ways. A big problem (though not the only one) with the traditional wording of both Newton's second and third laws is that they incorrectly suggest cause and effect (and hence imply a chain of events, as you ... 4 The flaw is that you've failed to do an experiment which will tell you whether the frame is inertial or not. If you do such an experiment -- for instance take a test mass, initially at rest with respect to the frame, release it, and see if it remains at rest -- you will immediately discover that the frame is not inertial. 0 This is the answer for one of previous question, It should be "Every action has equal and opposite reaction of SAME TYPE" , the best way to understand these contact force related questions is to draw a large clear free body diagram. Which will eventually lights up your problems. Draw all possible forces on both object(s) and or contact surface. Pls see ... 1 How about considering a specific force, such as the Newtonian gravitational force between two point masses m_1 and m_2. We could write the force that mass 1 exerts on mass 2 as follows:$$\vec{F}_{12} =-G\frac{m_1 m_2} {r_{12}^2} \hat{r}_{12},$$where$r_{12}$and$\hat{r}_{12}\$ are the distance 2 is from 1 and unit vector from 1 to 2, respectively. ...

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