Two bars of masses m1 and m2 connected by a nondeformed light spring rest on a horizontal plane. The coefficient of friction between the bars and the surface is equal to k. What minimum constant force has to be applied in the horizontal direction to the bar of mass m1 in order to shift the other bar?
My approach was to set the force on m2 = 0, so the spring force equals the friction force. Then I set the force on m1 = 0, so the force equals the spring force plus the friction force on m1. This gives the answer as F=k(m1+m2)g.
However the correct solution is found by setting the work done by block 1 = 0, which gives k(m1+(m2)/2)g.
Can someone help me physically understand why the solution would be found by setting the work done equal to zero rather than the net forces equal to zero?
Thanks for the help!
Two masses connected by a spring
Re: Two masses connected by a spring
The main problem in your solution, I think, comes from your neglect of the speed of $$m_1$$ when a force $$$F$$$ acts on it (infect, I did the same mistake when I solve this at first). If the bars were connected by a light cord instead of a spring, your answer is correct.
So,note that, when $$m_2$$ is fix, and $$$Fkm_1g>0$$$, under the influence of the spring, $$m_1$$ moves like a damped harmonic oscillator. The amplitude of the firsthalf period (from left to right, for example) satisfies:
$$$Fkm_1g=bA$$$,
where, $$$b$$$ is the spring constant. It is clear to see that the spring will be extended as long as $$$2A$$$ in this process, which leads the maximum force, exerts on $$m_2$$, $$$f=2bA$$$.
And thus we have:
$$2bA\leq km_2g$$
So, the minimum value of $$$F$$$, is given as:
$$$2(Fkm_1g)=km_2g$$$
namely,
$$$F=k(m_1+\frac{m_2}{2})g$$$
So,note that, when $$m_2$$ is fix, and $$$Fkm_1g>0$$$, under the influence of the spring, $$m_1$$ moves like a damped harmonic oscillator. The amplitude of the firsthalf period (from left to right, for example) satisfies:
$$$Fkm_1g=bA$$$,
where, $$$b$$$ is the spring constant. It is clear to see that the spring will be extended as long as $$$2A$$$ in this process, which leads the maximum force, exerts on $$m_2$$, $$$f=2bA$$$.
And thus we have:
$$2bA\leq km_2g$$
So, the minimum value of $$$F$$$, is given as:
$$$2(Fkm_1g)=km_2g$$$
namely,
$$$F=k(m_1+\frac{m_2}{2})g$$$

 Posts: 4
 Joined: Fri Mar 16, 2012 10:28 am
Re: Two masses connected by a spring
a slightly different method can be like this :
as the force is applied on the mass m1. and second block will move if the force exceeds um2g. this force will be applied by the spring. so minimum compression in the spring to provide this force is kx. so kx = um2g. where k= spring constant, u=friction coefficient
now apply energy conservation Fx  1/2kx^2 = um1gx
i.e. F = 1/2kx+um1g = um2g/2 +um1g
as the force is applied on the mass m1. and second block will move if the force exceeds um2g. this force will be applied by the spring. so minimum compression in the spring to provide this force is kx. so kx = um2g. where k= spring constant, u=friction coefficient
now apply energy conservation Fx  1/2kx^2 = um1gx
i.e. F = 1/2kx+um1g = um2g/2 +um1g