Two masses Definition and 241 Threads

For this problem, Why is the tension on each side not equal? For this problem I think the only assumption is that the string is inextensible so the accelerations of the masses are equal. Many thanks!

The doubt is about this question. Which the answer is They say that all the forces are pulling forces? How is this the case? How would this setup even work? I don't think block 2 could pull the rod.

I'm having some trouble figuring this problem out. I've found the tension in (a) but I don't know where to start with (b). I've found that the distance between one of the masses and the rotational axis on the picture is R+0.52 m and that the masses rise to a height of h = 0.3 m. The moment of...

We can find the potential energy by finding the potential difference between the two masses. the minimum distance between the two masses is 10 cm. The maximum is 30 cm because they can be 3 string lengths apart as they repulse each other once the string is cut. So, to get potential difference...

I have no clue how to do the problem. I created a free body diagram for each block. I assume that it is simpler than I am making it out to be.

Two bodies with masses m2 and m1, m2=a.m1 and m2>m1 orbit each other in circular orbits. if the barycenter moves at a speed v with respect to an inertial reference frame, What should be the relative speed of the bodies so that m1’s orbit in this inertial frame will have cusps?

Density of the Sphere = 3M/4πR³ Mass of carved out sphere = density × 4π/3 × R³/8 = M/8 The position of center of mass of The Sphere {M(0) - M/8(R/2)}/M-M/8 -R/14 So total distance between centers of the two bodies is R/14 + 3R = 43R/14 So now I found force between the Mass 7M/8 (left out...

1) By conservation of linear momentum: ##m_1 v_1-m_2v_2=(m+m_1+m_2)v_{cm}\Rightarrow v_{cm}=\frac{m_1}{m+m_1+m_2}v_1-\frac{m_2}{m+m_1+m_2}v_2=\frac{3}{8}\frac{m}{s}##; 2) By conservation of angular momentum: ##-Rm_1v_1-Rm_2v_2=I_{total}\omega=(I_{disk}+m_1R^2+m_2R^2)\omega## so...

I am just confused on how to find the normal force/ FN of the first object. My classmates are saying Fgy is the exact same as Fn but I don’t get why Fgy= Fg sin theta Fgy= (20)(9.81) (sin35) Fgy= 112.5 Fgy = FN

Figure : Option of question : Solution attempt :

I got (a) but have no idea about b. Potential fields aren't just additive all the time are they?

So far, I have found g of the foreign planet, Vf of the ball, and realize that GPEiA = KEfA (Am I right here?). Thus, since GPEiA = GPEiE, GPEiE also equals KEfE. I also understand that, since the same impulse is applied to catch the objects, both objects have the same momentum at the moment...

[Mentor Note: thread split off from a different thread] https://www.physicsforums.com/threads/heavier-objects-fall-faster.1002022/ Since seeing this thread yesterday, I have been trying to derive the time equation for the collision of two masses due to Newtonian gravity. Unfortunately, this...

i) I first analyzed the forces as soon as the 14 Kg is released. The aim of this step is to calculate the work done by the net force acting on the 14 kg mass to determine the change in kinetic energy. T-14g=-14a T-8g=8a T=99.7 N a=6g/22 m/s^2 Since the net force is constant and does not vary...

Hello everyone. Probably this question is trivial, but nevertheless I am confused about Newtons law of motion: $$F=G\frac{m_1m_2}{r^2}$$ Now, some sources say, that F is the force between the two masses m1 and m2. Other sources say, that F is the force that m1 exhibits on m2. But isn’t this a...

i attempted this problem by using conservation of energy, mgh=1/2kx^2 mgD=1/2kD^2 2mg=kD k=2mg/D why is it wrong ? btw the correct answer used mg = kx which is mg/D

So first I looked at the forces acting on m1 m1a1=F spring on m2-F spring Then m2 m2a2=F spring-F spring on m1 using 3rd law, m2a2=F spring+F spring on m2 m2a2=F spring+m1a1+F spring m2a2=2F spring+m1a1 Not entirely sure if I've done the above correctly, but I am stuck now because I have...

Answer is 7.00. I don't really know where to start on this one. Because there is no friction, shouldn't there be no force pushing against the first mass? If I assume the next mass has a force that acts against the first, I'm not sure how to find it.

hey there, I am confusing what is the right answer to this system of mass, I solved it a couple of time and I got different answers. can you help me with that? I need to find the expression of the acceleration of the 2 masses. thanks a lot,Adam.

Note: the working (taken from iWTSE website) refers to inertia as the symbol ‘J’ (in-case there was any confusion).I found equations of motion for mass m and 2m which were ‘T1 = ma + mg’ and ‘T2 = 2mg – 2ma’, respectively. I know they are connected particles with the same acceleration ‘a’.I have...

I think I have done this in an inefficient way. I assumed from the question that the three balances were of equal mass, though I'm not sure if I should have ignored their mass entirely. The correct answer is X = (62 + [2/9]) kg. Please help.

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I solved for T on m1 and arrived at 6.72. I plugged that value into the ΣFx equation as shown above (pardon my handwriting) and got a mass of 0.88 kg. The online program indicated that I needed to check my expression for tension, noting that the two tensions are heading in opposite directions...

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...

1) I found: ##x_{CM} = z_{CM} = 0 ## ##y_{CM} = \frac{L}{2} \frac{m_1 - m_2}{M + m_1 + m_2} ## 2) Applying the conservation of linear momentum: ##M v_0 - m_1 v_0 - m_2 v_0 = (M + m_1 + m_2) v_f ## ##v_f = \frac{M - m_1 - m_2 }{M + m_1 + m_2} v_0 ## The velocity should have been found also...

I tried in this way: 1) Considering a reference axis oriented downwards: ##M_1:## ## -T + M_1g - k \frac{L_0}{2} = 0## ##T = M_1g - k \frac{L_0}{2} ## 2) ##M_2:## ##M_2g + k \frac{L_0}{2} = M_2 a## ## a = g + \frac{k}{M_2} \frac{L_0}{2} ## 3) ## M_2g \frac{L_0}{2} + \frac{1}{2} k...

If M moves ##x## along the plane, her height variation in ##x \cos(\alpha)##, and, but I don't know how to find the variation of the height of ##m##

Hi! I need help with this problem. m1-2-3-4 and R are given. There is no slip in the system. I have to give F1-2-3-4 in respect of the masses and R. Here is what I managed to m1 is easy: m1*a = m1*g - T(tension of the rope) m2: m2*a = T - (?) <-- I have a problem with this. F1 and F3 is the...

A penny of mass 3.10 g rests on a small 20.0-g block supported by a spinning disk. The block is sitting at the edge of the disc at a radius of 12 cm. If the coefficient of friction between block and disk are 0.750 (static) and 0.640 (kinetic) while those for the penny and block are 0.450...

the entropy change for a reversible adiabatic process is zero as it remains constant. Is this a reversible process? assuming T1>T2: hot (h) water has mass M, temp T1 cold (c) water has mass nM, temp T2 let the final temperature be Tf if δQ=0 as the process is adiabatic, |Qh|=|Qc| so Qh=-Qc...

As ##P=ccte## we can find final velocity considering a plastic collision ##m_0 . v_0 =(2m+m_0).V## But what about the angular velocity? Because, as the bullet hits the centre of mass of the string, it won't have angular velocity

I really want to know which answer is correct. I don’t really know if I should include velocities to the left as negative velocities in the equation. Is it -1 or 4.33? Please help! Thanks!

Morning all I've recently come across a problem where I get conceptually but cannot apply mathematically if that makes sense. I understand the position of the third mass must be at the equilibrium point of ##m_1## (##9.0×10^{24}kg##), so ##\Sigma F = 0## right? And not even necessarily zero...

I want to know if my solution is correct: The velocity of the center of mass immidiately after m1 has moved is just vcm=m1v1 /(m1+m2) I thing that i have to consider the spring and the m2

I first tried to calculate the moment of inertia of the bar. The problem is that I don't understand exactly how are the dimensions of the bar. The fact that it has a width ##a## means that its height is ##a##, or that it has an unknown hight and the width is ##a##, like a parallelepiped? After...

Homework Statement Homework Equations F = ma The Attempt at a Solution I couldn't draw the freebody diagram. There's this weight of mass m, which is mg, downwards from m and there's tension T to upwards. This T affects mass M in the +x direction, but how could i find out the normal force...

Homework Statement Two identical masses are connected with a rope and are gliding without any friction. Situation given in the picture: Determine after which distance "s" they stop if we have s=0 at t=0 with starting velocity ##v_0## Given: ##\alpha## and ##\beta## with ##\alpha < \beta##...

Can anyone please explain to me how can I calculate the "Tension" and "acceleration" of M in this question interms of M,m1,m2 and g? I can't understand how M has an acceleration or why M is involved in the calculation of the tension my solution was that acc of M is zero and Tension = m1g x...

Homework Statement Two blocks are attached to opposite ends of a massless rope that goes over a massless, frictionless, stationary pulley. One of the blocks w/ a mass of 1.5 kg accelerates downward at (3/4)g. A. What is the mass of the other block? Homework Equations There were no given...

Homework Statement Two blocks of the masses m1=7.40 kg and m2=m1/2 are connected via a massless pulley and massless string. The system is currently in equilibrium but is about to start sliding, if m2 would increase even by a bit. For the friction between the surface and m1 assume that µs=µk...

1. A weightless rod carries towards of masses M and M. The roads Hinge Joint to vertical axis OO', which rotates with an angular velocity ω. Determine the angle φ formed by the rod and there vertical. The attempt at a solution If I am not wrong, the two ways to ensure equilibrium are...

Hello ! Homework Statement Two blocks M1 and M2 are connected by a string of negligible mass. If the system is released from rest, find how far block M1 slide in time ##t##. Neglect friction. Diagram: See Attached Image Clue given in the manual: If M1 = M2, then solution is ##x(t)=...

Homework Statement Statement of the problem (quoting from my assignment): a) write equations of motion b) try to solve analytically Given: m1, m2 - two masses R - distance between two masses Homework Equations V=-G(m1/r + m2/(R-r)) F=-dV/dr The Attempt at a Solution a) Equations of motion: v...

I have the following system I got from a book, which models the dynamics of two masses connected by a shaft. The system is given below: And the equations given in the book for this system is below: The nominal speed is w0. And the interest here is the deviations from the nominal for mass...

Homework Statement Take the x-axis to be pointing perpendicularly upwards. Mass ##m_1## slides freely along the x-axis. Mass ##m_2## slides freely along the y-axis. The masses are connected by a spring, with spring constant ##k## and relaxed length ##l_0##. The whole system rotates with...

Homework Statement [/B]Homework Equations F = m*aThe Attempt at a Solution [/B] Hello, i think i understand d and e so only the first three tasks are important.a) a = (m2 - m1)/(m1 + m2) *g -> (5kg - 2kg)/(2kg+5kg) *10ms^-2 = 4.29 ms^-2 T = m1 (a+g) -> 2kg (4.29ms^-2 + 10ms^-2) = 28...

Homework Statement The given system is released from rest. Assuming no friction, mass-less pulleys and ideal strings; calculate the accelerations of the pulleys. Homework Equations Constraint equations. F=ma The Attempt at a Solution Taking the above assumptions; And applying constraints...

Homework Statement Homework EquationsThe Attempt at a Solution i) ##x_{cm} = \dfrac{m_1x_1 + m_2x_2}{m_1 + m_2}## At any time t , ##x_{cm} = v_0t## and ##x_1= v_0t - A(1-\cos{ωt})## From the above two , we get ##x_2 = v_0t + \dfrac{m_1}{m_2}A(1-\cos{wt})## I'm not clear what to do in...