Finding the accel of two objects on an incline.

[Agent M27]

Homework Statement

Two objects are connected via a light string that passes over a frictionless pulley. Assuming the incline is frictionless, m₁=2 Kg m₂=6 Kg and \theta=55. Find the acceleration of the objects and the tension in string.

This porblem is set up with m₁ hanging vertically from the pulley and m₂ is resting on the inclined portion connect to the same string.

Homework Equations

\SigmaF=ma

The Attempt at a Solution

What I did is I found the tension in the string using the equation of motion for the mass m<sub>1/SUB]:

\SigmaF=T-mg=ma

T=m1g+m1a

I choose the acceleration to be positive here because were this a true situation this would be the direction of motion of the system.

I then wrote an equation of motion for the mass m2/SUB]:

\SigmaF= mgsin\theta-T=-ma

I choose this as the EQ of motion because mgsin\theta will provide more of a force than the tension of the rope, and I choose -ma because this portion of the system is traveling in the opposite direction.

Plugging in my equation for T I arrive at the following eq for the acceleration of the system, which is equal due to them being connected.

a=\frac{m_{2}gsin\theta - m_{1}g}{(m_{1}-m_{2})}

Basically I am wondering if I am correct, and also what is a solid method for determining the order of the terms in the EQ of motion? Thanks

Joe</sub>

 

[kuruman]

You are correct. I am not sure what you mean by "the order of the terms in the EQ of motion". You have a general equation that is always true. If, when you plug in the numbers, the acceleration turns out to be a positive number, then you have chosen its direction correctly. Else, the acceleration must be in the opposite direction from the one you chose.

 

[PhanthomJay]

Homework Statement

Two objects are connected via a light string that passes over a frictionless pulley. Assuming the incline is frictionless, m₁=2 Kg m₂=6 Kg and \theta=55. Find the acceleration of the objects and the tension in string.

This porblem is set up with m₁ hanging vertically from the pulley and m₂ is resting on the inclined portion connect to the same string.

Homework Equations

\SigmaF=ma

The Attempt at a Solution

What I did is I found the tension in the string using the equation of motion for the mass m<sub>1/SUB]:

\SigmaF=T-mg=ma

T=m1g+m1a

I choose the acceleration to be positive here because were this a true situation this would be the direction of motion of the system.</sub>

Your equation is correct, but since you chose T as positive, T>m_1(g), then a is positive, in the direction of the net force.

I then wrote an equation of motion for the mass m<sub>2/SUB]:

\SigmaF= mgsin\theta-T=-ma

I choose this as the EQ of motion because mgsin\theta will provide more of a force than the tension of the rope, and I choose -ma because this portion of the system is traveling in the opposite direction.</sub>

You have a signage error.

Plugging in my equation for T I arrive at the following eq for the acceleration of the system, which is equal due to them being connected.

a=\frac{m_{2}gsin\theta - m_{1}g}{(m_{1}-m_{2})}

Basically I am wondering if I am correct, and also what is a solid method for determining the order of the terms in the EQ of motion? Thanks

Joe

Correct your signage error. The acceleration of each block is in the direction of the net force. If m_1 is assumed to move up, as a positive direction, then m_2 moves down the plane, as a positive direction. If the net force on block is assumed down the plane as positive, then 'a' is positive down the plane. If 'a' comes out negative, you assumed the wrong direction for the net force.

 

[kuruman]

Oops. PhantomJay is correct. I looked at the numerator of your expression without paying too much attention to the denominator. If m₁ = m₂, the acceleration becomes infinite which is clearly troublesome.

 

[Agent M27]

So acording to your directives PhantomJay, my second equation of motion ought to be:

\SigmaF = m₁gsin\theta - T = m₂a

Plugging in T to solve for a I get the following:

a=\frac{m_{2}gsin\theta - m_{1}g}{(m_{2}+m_{1})}

This is the case because I choose the vertical component of gravity to be the positive, therefore net force, as well the accel is in the direction of this force, which made it positive. Thanks gents.

Joe