How Do You Calculate Forces and Accelerations in a Two Block and Pulley System?

[Metamorphose]

1. A block with mass m₁ is placed on an inclined plane with slope angle α and is connected to a second hinging block with mass m₂> m₁ by a massless cord passing over a small frictionless pulley. The coefficient of kinectic freiction between mass m_a and the incline is negligible. Find:

(a) the acceleration vector of mass m₁

(b) the acceleration vector of mass m₂

(c) the magnitude of the normal force on mass m₁

(d) the magnitude of the tension in the cord

[/b]

Homework Equations

Newton's Equations

The Attempt at a Solution

(a.) Drawing a FBD diagram for mass 1 gives that the sum of the forces for the x-component F_x = T - wsinα.

F_y = n - wcosα.

Since there is no acceleration in the y-direction, the only acceleration we need to worry about is that for the x-component.

ƩF_x = m₁a

m₁a = T - wsinα

a = (T - wsinα)/m₁

a = (T - m₁gsinα)/m₁?

I am not sure if this is correct or wrong. Please help me out here

 

[Doc Al]

So far, so good. But you're not done. You need to express the answer in terms of the given quantities only, not in terms of T. (Hint: Set up an equivalent equation for m₂.)

 

[Metamorphose]

An equation for m₂:

ƩF_y = m₂a_y

∴m₂a_y = T - m₂g

Solving for a in this case gives (T - m_ag)/m₂.

If we could find an equivalent for T in terms of m₂, we could plug it into the equation for part A. Can this be done by manipulating the same equation to obtain an equation in terms of m₂, a_y and g?

 

[Doc Al]

An equation for m₂:

ƩF_y = m₂a_y

∴m₂a_y = T - m₂g

Solving for a in this case gives (T - m_ag)/m₂.

OK, but use the same letter for the magnitude of the acceleration. And be careful with signs.

You'll have two equations (one for each mass) and two unknowns (the acceleration and the tension), which you can solve together.

 

[asteriatic]

.

(b) For mass 2, the FBD diagram gives that the sum of the forces for the x-component is Fx = m2gsinα - T.

ƩFx = m2a

m2a = m2gsinα - T

a = (m2gsinα - T)/m2

a = (m2gsinα - m1gsinα)/m2?

Again, not sure if this is correct or wrong. Please help me out here.

(c) The normal force on mass m1 is equal to the weight of the mass, which is m1g.

(d) The tension in the cord is equal to the weight of mass 2, which is m2g.


Your solution for part (a) is correct. For part (b), the correct equation is:

m2a = m2gsinα - T

Therefore, the acceleration of mass m2 is:

a = (m2gsinα - T)/m2

Your solution for part (c) and (d) is also correct. Good job!