Solving Mass-Acceleration Relationships for Objects Connected with Pulleys

[Geminiforce]

Homework Statement

An object of mass m1 on a frictionless horizontal table is connected to an object of mass m2 through a very light pulley P1 and a light fixed pulley P2 as shown below.
(a) If a1 and a2 are the accelerations of m1 and m2, respectively, what is the relation between these accelerations? (Use a_2 for a2, m_1 for m1, and m_2 for m2 as appropriate.)
(b) Express the tensions in the strings in terms of g and the masses m1 and m2. (Use g, m_1 for m1, and m_2 for m2 as appropriate.)
(c) Express the accelerations a1 and a2 in terms of g and the masses m1 and m2. (Use g, m_1 for m1, and m_2 for m2 as appropriate.)
http://www.webassign.net/serpop/p4-38.gif

Homework Equations

F=ma

The Attempt at a Solution

First I know that Tension2 = 2Tension1 due to mechanical advantage.
Therefore,
Tension 1= m1a1
Tension 2= m2a2
if i substitute 2T1 into T2, i get
2m1a1=m2a2 => a1=m2a2/2m1
The web assign says it is wrong. Any ideas on how to set this up?

 

[Doc Al]

First I know that Tension2 = 2Tension1 due to mechanical advantage.

Good.

Therefore,
Tension 1= m1a1

Good.

Tension 2= m2a2

Not true. (Tension is not the only force acting on m2.)

Set up Newton's 2nd law for m2.

You'll also need to figure out how a1 and a2 are related. (They are related, since the pulleys are connected by strings.)

 

[thomate1]

I would approach this problem by first analyzing the forces acting on each object. For object m1, there is a force of tension acting to the right and a force of gravity acting downward. For object m2, there is a force of tension acting to the left and a force of gravity acting downward. Since the pulleys are light and frictionless, we can assume that there are no other external forces acting on the system.

(a) Using Newton's second law, we can set up the following equations for each object:

m1a1 = T1 - m1g
m2a2 = m2g - T2

Since the two objects are connected by the same string, the tensions T1 and T2 are equal. Therefore, we can set T1 = T2 = T. Substituting this into the equations above, we get:

m1a1 = T - m1g
m2a2 = m2g - T

Solving for T in the first equation and substituting into the second equation, we get:

m2a2 = m2g - (m1a1 + m1g)

Rearranging this equation, we get:

m2a2 = m2g - m1a1 - m1g

Now, we can substitute the value of a1 from the first equation into this equation:

m2a2 = m2g - (m2a2/2m1) - m1g

Solving for a2, we get:

a2 = (2m1g - m1a1)/m2

Therefore, the relationship between the accelerations is a2 = (2m1g - m1a1)/m2.

(b) To find the tensions in the strings, we can use the equation T = m1a1 = m2a2. Substituting the values of a1 and a2 from the equations above, we get:

T = m1(2m1g - m1a1)/m2 = m2(2m1g - m1a1)/m2

Simplifying this equation, we get:

T = 2m1g - m1a1 = m2g - m2a2

Therefore, the tensions in the strings are T = 2m1g - m1a1 and T = m2g -

 

[baba89]

I would start by identifying the forces acting on each object in the system. For m1, there is the tension force from the string on one side and the normal force from the table on the other side. For m2, there is the tension force from the string on one side and the gravitational force pulling it down on the other side.

Next, I would use Newton's second law, F=ma, to set up equations for each object, taking into account the direction of the forces and the mass of each object. For m1, the equation would be T1 - N = m1a1, and for m2, the equation would be T2 - m2g = m2a2.

Since the pulleys are light and frictionless, we can assume that the tension forces are the same on both sides of each pulley, so T1 = T2 and N = m1g. Substituting these into the equations, we get T1 - m1g = m1a1 and T1 - m2g = m2a2.

Solving these two equations simultaneously, we can find the relationship between the accelerations a1 and a2: a1 = (2m2 - m1)g / (2m1 + m2).

To find the tensions in the strings, we can substitute this relationship into the equations for T1 and T2: T1 = m1a1 = m1(2m2 - m1)g / (2m1 + m2) and T2 = m2a2 = m2(2m2 - m1)g / (2m1 + m2).

Finally, to express the accelerations in terms of g and the masses, we can substitute the expressions for T1 and T2 into the equations for a1 and a2: a1 = T1 / m1 = (2m2 - m1)g / (2m1 + m2) / m1 and a2 = T2 / m2 = (2m2 - m1)g / (2m1 + m2) / m2.

Overall, the relationship between the accelerations is a1 = (2m2 - m1) / (2m1 + m2) * a2, and the tensions in the strings are T1 = (2m2 -