What is the Acceleration of a Mass on an Incline Plane?

[Koscher]

Homework Statement

A hanging mass, M1 = 0.493 kg, is attached by a light string that runs over a friction-less pulley to a mass M2 =1.81 kg that is initially at rest on a friction-less ramp. The ramp is at angle 30.5 above horizontal. Find the magnitude and direction of the acceleration, a2, of M2.

Homework Equations

F=ma
T-mgsin(theta)=ma
F(perpendicular)=M*cos(theta)
F(parallel)=M*sin(theta)

The Attempt at a Solution

(.493)(9.81)-(1.81)(9.81)sin(30.5) = 1.81a
a = -2.306 m/s^2

But that is not the right answer. So I am lost.

 

[Crwth]

Are you sure about the sign for the acceleration?

Which direction is defined as positive or negative?

 

[Koscher]

Take a positive result to be "up the ramp" and a negative result to be "down the ramp".

That is what the help my professor gave the class about the signs. Which i believe that it is a negative sign but i could be wrong.

 

[gneill]

Homework Statement

A hanging mass, M1 = 0.493 kg, is attached by a light string that runs over a friction-less pulley to a mass M2 =1.81 kg that is initially at rest on a friction-less ramp. The ramp is at angle 30.5 above horizontal. Find the magnitude and direction of the acceleration, a2, of M2.

Homework Equations

F=ma
T-mgsin(theta)=ma
F(perpendicular)=M*cos(theta)
F(parallel)=M*sin(theta)

The Attempt at a Solution

(.493)(9.81)-(1.81)(9.81)sin(30.5) = 1.81a
a = -2.306 m/s^2

But that is not the right answer. So I am lost.

It looks as though you tried to apply your formula T-mgsin(theta)=ma but used M1*g for the tension, T. This isn't correct. When the masses are allowed to accelerate the tension in the string will be less than that.

If you write that formula separately for each of the blocks leaving T as an unknown, you'll have two equations in two unknowns (a and T). Eliminate T and solve for a. Note that for the first block, because there is no slope (or you could say that the angle is 90° for it) the formula simplifies to: T - M1*g = M1*a . Be sure to pick a direction for positive acceleration and adjust the expressions accordingly (you can multiply the LHS by -1 to change the sign).

 

[Koscher]

Thank you, that makes sense, I solved it with T as an unknown and got the right answer.