Newton 2nd law Definition and 142 Threads

The first question statement was under the chapter ##Newton's Laws Of Motion (Without Friction)##. Whereas, the second question was under ##Friction##. The free body diagram for the first question is given as: And the free body diagram for the other question is given as: In the first...

Body M has a mass of 80kg and monkey has a mass of 20kg. No friction between table and the object M. Can someone please explain to me if the acceleration of the system will be greater with monkey accelerating up or down(relative to the rope), and why. Monkey accelerates both up and down at...

Is there some kind of rule when we are supposed to use Fn×mu compared to F×mu or mg×mu to calculate friction? Because I don't get the same answer when I use Fn×mu and F×mu. In magnitude the forces are the same but they are not same in directions. Why is it so? Here is the photo. Problem goes...

I wonder why did I get two different answers for m2 using two different approaches? The correct answer is m1=3kg, m2=7kg. So pleases if someone can take a look. Also can someone suggest how am I supposed to find the force that dynamometer is showing? I was thinking that it equals (m1+m2)*g...

Apparently body A has greater acceleration and I don't get why. Can someone please explain. It's stated to take g=10/s^2. Both bodies have same mass. Nothing about rope elasticity is added, and nothing about the pulley. So I assumed they are both massless, and no friction between them. Also...

I've tried to solve this exercise but I haven't used one of the properties of the system (the displacement of the masses) so I don't know if I'm wrong about my procedure. First of all, we (obviously) know that $$ P=P $$ And since we can express the power of a force in two different ways, we...

I have a doubt with the last part. Why isn't the right side negative? Because when the block is released, it moves upwards. thanks image was obtained from here Thanks.

In order to calculate the acceleration, I sum the equations sum fx^B and sum fx^A. F_{r,B} - m_Ag sin(theta) - (m_A+m_B) g sin(theta) = - (m_A+m_B) a then a = -[F_{r,B} - (2 m_A+m_B) g sin(theta)]/(m_A+m_B) =-[muk (m_A+m_B) g cos(theta) - (2 m_A+m_B) g sin(theta)]/(m_A+m_B) but the...

I don't undertand the equation. It is Newtons's second law of motion, so it decribes a force that acts on a single disc relative to the ground. So when the force is proportional to velocity, shouldn't it be ##-bv##? Because the dics's velocity is ##v## relative to the ground. Relative to the...

Hi, I am having trouble with this problem. I'm thinking the solution is this but I'm not sure. Fnet=m1a+m2aFnet=m1a+m2a , m1a=kxm1a=kx, m2a=F−kxm2a=F−kx so x=m1ak=−(m2a−F)kx=m1ak=−(m2a−F)k

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The mathematical representation of the net external force on a system(obtained from Newton's second law) is ##\vec F_{net} = \frac {d\vec P}{dt}##, which is the rate of change of linear momentum of the system. If we substitute ##\vec P = m\vec v## into the formula for force and differentiate, we...

At the beginning, I just looked for the highest point in the graph, which is approximately 90 \frac m s^2. Then I plugged it in the formula F = ma and got the force equaled to 0.045 N. However, when I looked back, the graph is about the change in acceleration. So really, I'm dealing with...

So i got some equations but i think i am missing something, my main doubt is what is the relation between dx / dt and v(o) [ here] . Workings in attachment

When I exert enough force that overcome the static friction. The object start moving and surface create kinetic friction on object if I exert harder overcome the maximum of friction it start accelerate. When I release the object will the kinetic friction disappear immediately proportional to...

I'm stuck to understand 3 laws of Newton. It doesn't make sense to me. 1. Suppose the case when a person stay in a rest vehicle.When we accelerate the car the person still at rest so the person has to move to the tail of the vehicle cause his intertia of staying rest. If we decrese the...

Assume that the lengths of various sections of the string at an instant are as shown. So we can say that ##l_1 + l_2+l_3= k## at an instant of time ##t##, where ##k## is a constant. Taking the second derivative of both sides w.r.t. ##t##, we get $$\frac {d^2{l_1}} {dt^2} +\frac {d^2{l_2}} {dt^2}...

This is a problem that was posted here in 2003 and is now closed for replies. This question can be found at https://www.physicsforums.com/threads/friction-problem.662/ The answer in that old post didn't seem clear to me probably because it was highly summarized. There was no mention of static...

I tried splitting the forces up into F1 and F2 making Newtons second law equation into F1+F2=ma. Then I added over the the first force given. multiply the mass to the acceleration terms to get F2= (m*ai + m*aj) - F1

I have used the work energy theorem like all source have shown me an have arrived at the right answer where work one by all the forces is the change in kinetic energy -1/2kx^2 - umgcosΘx +mgsinΘx = 0 is the equation which becomes -1/2kx -umgcosΘ+ mgsinΘ = 0 where k= spring constant u=...

Why F=m*a i.e product or multiplication and not F=m+a? i.e addition or summation?

So, to obtain the motion equations I initially plotted the free-body diagram (see picture). Then I’ve tried to get equations, but I’m not sure, do I have done it rightl. I will be gratefull if someone could help me.

In the solution for question ##(a)##, it is written that the equation of translational motion for the center of mass is ##N-mg=ma_y## Why ##N## is also included inside of the equation? In my opinion, the rail does not exerting force (N) to slow down the mass' acceleration. Instead, the purpose...

Hi everyone is able to help solve this question for my assignment in university? I've draw a free body diagram for each component of the question but now i am stuck. [Mentor Note -- Poster has been reminded to show their work when starting a new schoolwork thread]

I thought that the force by the pivot A on the pole AB would be the reaction force to the x-component of the gravitational force on AB. This would mean that the force by the pivot would be parallel to the pole, but in my notes from class the force vector seems to be more along the bisector of...

So I have a trolley of mass m that moves on a straight line. A sphere of mass m, is attached on the trolley with a light string of length a and it is left to oscillate. Just to give some idea of their positions: r_trolley = xi r_sphere = (x-asinθ)i - acosθj (θ is the angle between the string...

Here is where I am at : 1- F(C2/P) > F(C1/P) must be true I believe this is false, considering the angle separating C1 from the 'ceiling' is bigger 2- F(C2) > 10 N may be true I believe this is true because F(C2) could be broken down into vector components which may go above 10 N 3- Vector sum...

Forces: Box--> W(weight) and T(tension) Rope-->T1(reaction of T) and T2(because of the helicopter) So first i calculated Weight: W=mg=400*10=4000N In order to find the acceleration i should use Newton's 2nd law so: (Box) : T - W = ma T - 4000=400a The problem is with the rope...

Hi, I need to prove that the tension for artificial satellite consists of two points of mass m/2 connected by a light rigid rod of length , the tension in the rod is - $$ T=\frac{3}{4}\frac{Gmm'l}{a^3}-\frac{1}{4}\frac{Gm^2}{l^2} $$ the satellite is placed in a circular orbit of radius a>>l...

What does it mean that the relationship between material mass and weight is constant and proportional? Hi! Yes, another question... I have many doubts. :) I hope someone can help me with this apparently very basic doubt, but I feel like a stupid monkey trying to join two sticks to reach bananas...

Below is an ideal mass pulley system that we encounter in many problems under Newton's Second Law of Motion questions. Its supposed to be massless and frictionless i.e. string slips over the pulley and pulley does not rotate. In a real system, the pulley is assumed to be massless, whereas in...

Hi, I have to find the motion of a particles ##(x,y,z)##. However, I'm not sure where to begin. Is it correct to split the problem and first find what's the motion in the x direction then y and z. For exemple, ##m \frac{d^2x}{dt^2} = -kv_{0x} + qv_{0x}B sin 90 ## ##m\int\int...

I have trouble solving this problem any help would be appreciated.Problem statement ##J=\frac{mr^2}{2}## a) Determine the motion of yoyos for ##n=1,2,3## The case for ##n=1## is simple, however, I am having trouble with ##n=2## and ##n=3##. for ##n=2## I started by drawing all the forces...

At first I tried solving the problemteh following way: Due to symmetry let the rods connected to the green rod have tension forces in magnitde T1 => mg = 2T1cos(a), where a is half the angle formed by the two rods. From tere I got an expression from the longer rods in the force projected by them...

I did part a and part b but stuck in part c. Could you help me? (Part a and b attached.)

We have 2 forces affecting the rope: 1. Gravitational force of the body ##=mg## and 2. Force of air = Force of drag= ##F_{AIR}##. The length of the rope is shortening with the velocity ##v_k##. So to figure out the angle ##\theta## I wrote: ##R##= force of rope ##R_x = F_{AIR}## ##R_y = mg##...

From this question, I do not understand why there are three forces exerted at Point C (2 of it being the tension by weight A and the other is the tension by weight B) I understand that there is tension by the two weights but why is there 2 forces exerted by weight A at point C? From the...

As far as I understand it general relativity does not explain the origin of the inertial mass ##m_i## in Newton's law of motion ##\vec{F}=m_i\ d\vec{v}/dt## but rather it simply applies the concept to curved spacetime. For example if we have a particle with inertial mass ##m_i## and charge...

I have found the apparent weight of the man in the accelerating elevator My question is about what speed I should use in the formula for the power output P = F ⋅ speed My common sense tells me that the speed I should us is the speed v the man is climbing the ladder , because if the...

So I started by checking the options.I substituted the value of friction in the equations I got by making free body diagrams.I got different value of Tensions.For 1 and 3 Tension came to be 38 N.For 2 Tension came out to be 42N and for 4 Tension came out to be 40 N. Now I think that I will take...

I am not sure which other forces I should consider besides those 3. I cannot consider tensions due to the massless rod on the masses since those will not add up to zero.

Answer choices: N2L for Translation, N2L for Rotation, Both, Either1. You are asked to find the angular acceleration of a low-friction pulley with a given force exerted on it. My solution = N2L for rotation 2. You are asked to find the angular acceleration of a low-friction pulley due to a...

So ##T+U=\frac{1}{2}m(\dot{x}^{2}+\dot{y}^{2})-mgy=constant##. If I derive this with respect to ##t## $$\dot{x}\ddot{x}+\dot{y}\ddot{y}-g\dot{y}=0$$ Then I use ##\dot{y}=\dot{x}\frac{dy}{dx},\ddot{y}=\ddot{x}\frac{dy}{dx}+\dot{x}^{2}\frac{d^{2}y}{dx^{2}}## to get...

Let T be the tension in the string, a be the acceleration of mass 2m, 2a be the acceleration of mass m T = (m) (2a) ---eq(1) The mass 3m will come down with acceleration a’ = (a+2a)/2 = 3a/2 3mg - 2T = 3m . 3a/2 from equation 1 3mg - 2(2ma) = 3m . 3a/2 thus a = 6/17g thus acceleration of 3m...

I tried out the F=Δp/Δt equation and i came up with a change of momentum of the box-bullet fusion (M+m)V-0, but the textbook says otherwise (using a change of momentum of JUST the box itself, excluding the bullet). According to the textbook, the correct change is MV-0, without the added mass of...

I am writing about the nature of force in classical mechanics and what does really imply, in terms of change in motion. I am using as an example a circuit, on which we exert a force. I am trying to justify the following scheme (concretely, ##f_{mag}##): The thing is that I am wondering how...

On object 2: There are only 2 horizontal forces - Friction and Tension (of the spring). T = km2g On Object 1: There are 3 horizontal forces and the minimum value for F is when: F - km1g - km2g = 0 F = kg(m1 + m2) However, Solution is: F = kg(m1 + 0.5 m2) Any opinion?

Homework Statement This is the problem from the book "physics for scientists and engineers..." by Serway, Jewett / Chapter 5, problem 98 Initially, the system of objects shown in Figure P5.93 - is held motionless. The pulley and all surfaces and wheels are frictionless. Let the force F be...