Classical mechanics Definition and 1000 Threads

We know the minimum force to move the body in presence of friction will be ##F>\mu F_N## But it’s not right.

https://www.feynmanlectures.caltech.edu/I_09.html “Weight and inertia are proportional, and on the earth’s surface are often taken to be numerically equal, which causes a certain confusion to the student. On Mars, weights would be different but the amount of force needed to overcome inertia...

##T-2mg=2ma_1## (acceleration of heavier mass) ##T-mg=ma_2## (##-a_1=a_2##) On solving the eqns, ##a_1=-g/3=-a_2## ##s=1/2at^2## ##s=-g/6## , distance covered by heavier mass. ##s=g/6## , covered by lighter mass. Edit: ##\Delta U_1=mgh=-2mg^2/6## (decrease in U of heavier mass) ##\Delta...

##W_{ext}=mgh+KE_f+0=-20(5\sin 37)+(1/2)2(10^2)## ##W_{ext}=-60.18+100=39.81J## But it’s not consistent with ##W=F.d=20*5=100J## I can’t figure it out.

a) $$v=u+at$$ $$v=0+(F/m)t$$ $$v=-10$$ Now $$v^2=0+2as$$ $$s=-5$$ $$W=F.d=(-20)(-5)=100J>40$$ b) $$W=-mgh$$ $$W=-20(3)$$ $$W=-60$$

https://futurism.com/why-you-can-never-actually-touch-anything/amp We know the friction happens because of the intermolecular forces between the atoms of the surfaces in contact. When we place something on other there are places where they get come in contact there are bonding formed and due to...

For lower half ,$$Fnet=-\mu F_N+mg\sin \phi$$ For upper half, $$v^2=u^2+2as$$ (s is half of the total slant distance) $$v^2=0+2\frac{mg\sin \phi}ms$$ $$v=\sqrt{2g\sin \phi s}$$ again for lower half, $$v^2=u^2+2as$$ $$0=2g\sin \phi s+2\frac{-\mu F_N+mg\sin \phi}ms$$ $$\mu=\frac{2gm\sin...

As we always say eat food and rest because tomorrow you have to go to work. When we throw a ball up, it gains a max PE at some height. More the height more energy it has which makes it do more work. Is this right to say?

I think on top of the wedge the KE of both the wedge and block will be same but this fact doesn't take me anywhere. The base length of the wedge is not given. Maybe that would have helped.

I think if the two parts move in -x and +y direction, it must be balanced by the resultant of the two vectors but in opposite direction. So ##p\sqrt5## will be the answer. But I don’t think this is the right way to solve this.

##W=-k(y\hat i+ x\hat j)d##. I am not getting the coordinate of particle correctly so that I can find the value of d. Also the force is varying.

d=5m But the force is not mg since there is water so net force must be less. But how much?

The force is -Mg and distance is L/3. So W=-MgL/3. Not right! Maybe it’s done using COM but there is no additional information is given.

I have a question understanding the reasoning in the book. The book says in one dimension F=-dU/dr(p.185). From this, the system is stable at distance a when U'(a)=0 and U''(a)>0 where U is differentiated with respect to r.(p.217) My question arises from the instance of a pendulum where a...

W=-mgd=2*10*1.5=30J. I don’t think this is right.

We know that the force of gravitation is F=GMm/r^2. The acceleration of the body of any mass m is a=GM/r^2 which we call g. So same acceleration regardless of any weight(ignoring air resistance). But when we solve laws of motion problems we take a downward force of -mg. So force depend on mass...

I think because both are launched with same speed so both have same KE. Since one is Thrown downwards it’s KE will increase but not that much as the body which is thrown upwards. Because it covers more distance so it gains more energy(the body thrown up). So 3rd option must be right.

We know that the centripetal force and the tangential velocity is responsible for the motion of Earth around sun. Newton’s second law says F=ma. If we all get together(whole population) in one place like parallel to the tangent to orbit and jump can we displace Earth from its orbit?

When a tomato is thrown up with a velocity ##v_0## it’s kinetic energy is 1/2mv_0^2. It will stop at the top and then again comes back to the launch point where it’s kinetic energy will be same as before, 1/2mv_0^2. How is this possible? Also when we throw the tomato up how can you be so sure...

Its Good to be Back! From Resnik, Fundamentals of physics: Consider a particle of mass m, moving along an x-axis and acted on by a net force F(x) that is directed along that axis. The work done on the particle by this force as the particle moves from position ##x_i## to position ##x_f## is given...

Sometimes there are forces which act for very small time known as impulsive forces. We cannot measure such large forces acting in a very short time but we can ##\Delta p##. ##\Delta p=F\Delta t##.This quantity is defined as Impulse. Why do we keep introducing new quantities? When is a new...

Previously I have posted two threads on Thread 'Monkey climbing up the rope' https://www.physicsforums.com/threads/monkey-climbing-up-the-rope.1012065/ and Thread 'Car's maximum acceleration on a road is proportional to what?'...

What is going on when the ball is thrown up in the sky. It is pushed by a force F for some distance d. Then the object travels a distance s up in the sky before finally coming to a stop. So what is going on here? Is the force doing work for distance d or distance s (s>d)? I think change in...

1. From resnik, Halliday “Kinetic energy K is energy associated with the state of motion of an object. The faster the object moves , the greater is the kinetic energy” If I am right this means that greater the kinetic energy, greater is its speed. 2. Force transfers energy to the body due to...

Net force acting downwards is ##5g-g=4g## in downward direction. Net force on pulley is ##2T-4g##. Weight measured by balance?

##Net force=50-T2+T2-T1+T1-20=10a## ##a=3 m/s^2## I want to ask when we do this way rather than taking individual masses we can’t decide the direction of the system as we can when they are taken individually. So is it correct to just leave ##a## as it is and solve ?

I have gathered everything from post “Monkey climbing up the rope” about tension. Tension is basically a force that the rope applies back when it is under stress. It is an inward force. Tension T's direction at end points of rope where its attached to the body and ceiling is inwards. Tension is...

“Momentum is clearly a vector quantity. The following common experiences indicate the importance of this quantity for considering the effect of force on motion. 1. Suppose a light-weight vehicle (say a small car) and a heavy weight vehicle (say a loaded truck) are parked on a horizontal road. We...

https://www.feynmanlectures.caltech.edu/I_08.html Page 9-3. It says “In these terms, we see that Newton’s second law, in saying that the force is in same direction as the acceleration,is really three laws, in the sense that the components of force in the x, y,z directions is equal to the mass...

This problem is similar to what I have done before here. I think since the system is in equilibrium, that is both bodies are at rest, net force on each should be zero. So to balance the forces in all directions we need only friction forces on each in upward direction. So the force on B due to A...

Since no body accelerates so net force is zero. Force on each mass is zero. T1 and T2 both are 60N. Edit: since there is a force applied so there is acceleration on friction less surface.

The answer should be no change but we know ##F=ma##. In this eqn when acceleration increases mass decreases for same force. So why not here? If normal is doubled ##\mu## should be halved.

If I draw the fbd then some force will accelerate the car in horizontal direction which I think does not effect the string in vertical direction. So same tension regardless of acceleration. But we know it will increase. So what will be the correct physics behind it?

I think the tension in the rope will be equal to its weight , mg. I want to ask what if the monkey accelerates up with acceleration a, then what will be the tension in the rope?

1. ##-f_k\cos\theta-T\cos\theta+F_n\cos\alpha=m_2a_x## 2. ##f_k\sin\theta+T\sin\theta+F_n\sin\alpha-m_2g=-m_2a_y## 3. ##T-m_1g=m_1a_y## I am unable to get anywhere. There are accelerations in x , y directions. I need the value of acceleration. Then I can simply use ##s=ut+\frac12at^2##

We can differentiate twice the y displacement with respect to time t and get the acceleration of block B. $$a_B= \frac12 m/s^2$$. But I don’t think it’s that simple.

Since the car starts from rest it’s accelerating. So, $$F_a-f_k=ma$$ $$F_a-\mu mg=ma$$ $$\frac{F_a-\mu mg}m=a$$ Now from second eqn, ##s=ut+\frac12at^2## $$s=\frac12\frac{F_a-\mu mg}mt^2$$ $$\frac{2sm}{F_a-\mu mg}=t^2$$ $$\sqrt{\frac{2sm}{F_a-\mu mg}}=t$$ I don’t think I am getting any where!

I don’t know what is contact force. Are friction and normal forces called contact forces? And we have to take the resultant of the two to get the net contact force?

I think there is normal force from wall and applied force will balance each other but there is no counter force against mg.Both bodies will slip and fall. I am not sure.

$$a= -40/(10+6+4)$$ $$a=-2 m/s^2$$ Taking one mass of 10 kg. $$T-40=10(-2)$$ $$T=20 N$$ This is correct. But if I make the eqn of the system then $$-40+T-T+T-T=20(-2)$$ I have also drawn the diagram. It looks like the second body m2 is subject to no force. But it’s accelerating. How?

I read friction is the vector sum of the forces acting between the atoms of one surface and another. But the direction of friction is parallel to surface. I don’t get how friction occurs. I will also add the paragraph from my book “resnik” for further clarification. Other question is...

##(T-m1g)-(T-m2g)=(m1+m2)a## But I think this is wrong eqn.

I want to ask in this machine when there are two masses hanging down then the tension T is directed upwards along the rope. Is it the force applied by the rope on the mass? Is it the force applied by the pulley? When the anyone mass of the machine moves downwards is it because the force of...

##v^2=u^2+2as## ##v=10m/s## Again ##v^2=u^2+2as## ##a=25 m/s^2## So ##F=(200*25)/1000## ##F=5 N## Wrong answer!

Assuming it’s one body whose initial speed is u. First it attains height h then H. t1 and t2 are two times at which they attain h and H. ##h=ut1-\frac12gt1^2## ##H=ut2-\frac12gt2^2## ##\frac {t1}{t2}=1/3## Replacing t2 with 3t1, I am stuck.

##v=\frac12 *1*4+\frac12*1*4= 4 m/s## but the answer is wrong.

Integration of v= integration of##(alpha \sqrt x)dx##. But I am getting wrong answer.

3) and 4) is easy. Average and instantaneous acceleration is same from 1 to 4 sec since it’s constant acceleration. 1) and 2) I am unable to get the correct area under the curve. ##s = \frac 12*(-5)*1 + \frac12*10*2 + 10*1 + \frac12*10*1## Same I guess will be distance. 5) what is the graph...

I calculated v=0 at t=3. s(3) =24 m. s(5)=16 m. So reverse distance that the particle travelled=24-16=8 m. So total distance =24+8=32 m.

[Note: Link to the quote below has been pasted in by the Mentors -- please always provide attribution when quoting another source] https://www.feynmanlectures.caltech.edu/I_08.html Let s=16t^2 and we want to find speed at 5 sec. s = 16(5.001)2 = 16(25.010001) = 400.160016 ft. In the last...