Two Masses, a Pulley, and an Inclined Plane help

AI Thread Summary
The discussion focuses on solving the problem of two connected masses over a frictionless pulley, with one mass on an inclined plane experiencing kinetic friction. The equations of motion for both masses are established, leading to the expression for the mass ratio m1/m2. A key point raised is that a term was omitted in the initial solution, specifically the acceleration 'a' in the numerator. The correct ratio is derived as m1/m2 = (gsin(theta) + mu gcos(theta) + a) / (g - a). The conversation emphasizes the importance of including all relevant terms in the equations to arrive at the correct solution.
pcmarine
Messages
6
Reaction score
0
Block 1, of mass m1, is connected over an ideal (massless and frictionless) pulley to block 2, of mass m2, as shown. Assume that the blocks accelerate as shown with an acceleration of magnitude a and that the coefficient of kinetic friction between block 2 and the plane is mu.

MLD_2l_2_v2_2_a.jpg



Find the ratio of the masses m1/m2.

:confused:
 
Physics news on Phys.org
Let the tension in the thread be T. ok. now eq of motion of the 2 masses are...
(m1)g - T = (m1)a
T - (m2)gsin(theta) - (mu)(m2)gcos(theta) = (m2)a

Solving them by addng the two eqns u get,

m1/m2 = (g(sin(theta) + (mu)cos(theta)))/(g - a)
 
Thanks rammstein, but after inputting m1/m2, I get "aren't you missing a term?" Thanks a bunch for the help though!
 
pcmarine said:
Thanks rammstein, but after inputting m1/m2, I get "aren't you missing a term?" Thanks a bunch for the help though!

That's right. The equations of motion are correct, he just dropped a term when he solved for m1/m2. Go ahead and solve the system and see what you get.

-Dan
 
I tried setting the two equations equal to each other through T, but was unable to discover the missing term...
 
pcmarine said:
I tried setting the two equations equal to each other through T, but was unable to discover the missing term...

m_1g-T=m_1a So T=m_1g-m_1a

(m_1g-m_1a)-m_2gsin \theta-\mu m_2gcos \theta=m_2a

m_1g-m_1a=m_2gsin \theta+\mu m_2gcos \theta+m_2a

m_1(g-a)=m_2(gsin \theta+\mu gcos \theta+a)

\frac{m1}{m2}=\frac{gsin \theta+\mu gcos \theta+a}{g-a}

rammstein left out the last "a" in the numerator.

-Dan
 
  • Like
Likes Qexyario
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top