Objects sliding on inclines without friction (connected by a pulley and rope)

[PhysicS FAN]

Homework Statement

The 2 objects slid on the incline level without friction. The height difference of the objects at t=0 is 8m. At t'=1.5sec the objects are on the same horizontal level. What is the analogy of M/m? (M is the mass of the left object and m the mass of the right object.

*Have a look at the picture so that you can understand better the problem

Homework Equations

The Attempt at a Solution

I tried to use energy theorems, motion equations and also dynamics but I just can't get the right answer. Please help me !

 

[Delta2]

Your first step would be to find the constant common acceleration of the system. To that goal, apply Newton's second law to the body of mass M and then to the body of mass m . You will get two equations with two unknowns, the string tension and acceleration. Solve the system of two equations for acceleration.

After that you can use kinematics equations to find the point where they are at the same height after t=1.5sec has elapsed. You will get one more equation that will involve acceleration, the two angles 30,60 and the initial height difference d=8. You should be able to solve that equation to find the ratio m/M.

 

[PhysicS FAN]

Your first step would be to find the constant common acceleration of the system. To that goal, apply Newton's second law to the body of mass M and then to the body of mass m . You will get two equations. Solve the system of two equations for acceleration.

Yes that was what I tried to do first, but the whole system is connected with a thread that has an unknown tension. So the equation for M becomes:
M*g*sin60°- T = M*a (where T is the tension of the thread). Now because T is unknown I can't proceed with the system.

 

[Delta2]

You can proceed if you also write down the equation for Netwon's second law for the mass m. You ll get a second equation with also $T$ and $a$. You can combine the two equations and solve for acceleration $a$. (if you add the two equations, the Tension T will get simplified and will leave as only unknown the acceleration a).

 

[PhysicS FAN]

You can proceed if you also write down the equation for Netwon's second law for the mass m. You ll get a second equation with also $T$ and $a$. You can combine the two equations and solve for acceleration $a$. (if you add the two equations, the Tension T will get simplified and will leave as only unknown the acceleration a).

Equation for mass m: m*g*sin30°- T = m*a ⇒ a= (m*g*sin30 - T)/m and for mass M: a=(M*g*sin60 - T)/M so we combine the two and we get

m*g*sin30*M - T*M = M*g*sin60*m - T*m The tension can not be simplified. Besides we can find acceleration since it is given that after a known time the 2 objects are on the same horizontal level, which means that if the original height differnce was 8m the distance that both objects should cover is the half of the original. So we know the distance and the time and we can get acceleration. But that won't help either

 

[Delta2]

The correct equation after applying Newton's 2nd law for mass m is slightly different it is $T-mg\sin30=ma$. Solve now for acceleration a, you ll find an expression for acceleration that involves M,m,g and the two angles. Put in that expression $M=\rho m$ $\rho>1$ and do some algebra to express acceleration as a function of $\rho,g$ and the two angles.

 

[PhysicS FAN]

The correct equation after applying Newton's 2nd law for mass m is slightly different it is $T-mg\sin30=ma$. Solve now for acceleration a, you ll find an expression for acceleration that involves M,m,g and the two angles. Put in that expression $M=\rho m$ $\rho>1$ and do some algebra to express acceleration as a function of $\rho,g$ and the two angles.

Ok thank you I'll try that

 

[Delta2]

After you do that, notice than in time $t'=1.5sec$ both bodies move $s=\frac{1}{2}at'^2$ each along its own incline. But the height of mass M that moves down is $s\sin60$ while the height of mass m going up is $s\sin30$. you know that both heights add to 8m. So you ll get a third equation that involves the acceleration a. Put in this third equation the expression for acceleration you got from the previous step (where we simplify the tension). Solve this last equation for $\rho=\frac{M}{m}$.

 

[PhysicS FAN]

After you do that, notice than in time $t'=1.5sec$ both bodies move $s=\frac{1}{2}at'^2$ each along its own incline. But the height of mass M that moves down is $s\sin60$ while the height of mass m going up is $s\sin30$. you know that both heights add to 8m. So you ll get a third equation that involves the acceleration a. Put in this third equation the expression for acceleration you got from the previous step (where we simplify the tension). Solve this last equation for $\rho=\frac{M}{m}$.

Well sorry I'm asking again but I try and I can not simplify the tension

 

[Delta2]

ok, you have two equations
Newton's 2nd law for body of mass M : $Mg\sin60-T=Ma$
Newton's 2nd law for body of mass m : $T-mg\sin30=ma$

What do you get if you add these two equations?

 

[PhysicS FAN]

ok, you have two equations
Newton's 2nd law for body of mass M : $Mg\sin60-T=Ma$
Newton's 2nd law for body of mass m : $T-mg\sin30=ma$

What do you get if you add these two equations?

Yes you're right. Thank you. You've been really helpfull !