Calculating Acceleration and Tension in a Pulley System with Inclined Planes

[emma3001]

Please help me because my teacher has not explained how to do questions involving angles.

Blocks X and Y of masses mx=5.12kg and my=3.22kg are connected by a fishing line passing over a frictionless pulley. Show that block X slides up the incline (35.7 degrees above the horizontal) with positive acceleration. Determine the magnitude of the acceleration. (0.273m/s2 is the answer)

i want to find the gravitational force for mass x so Fg=5.12x9.8=50.2N
For mass y f=mg =3.22x9.8=31.6N

Just like with projectile problems you need to find the x and y components of Ftension for block x,which is getting pulled up at an angle but how do i do that if i don't know the applied force?

 

[learningphysics]

So the block y is hanging vertically?

Write the \Sigma F=ma equation for the block y, and the \Sigma F=ma equations for block x

Call the tension T. The acceleration of block y downward equals the acceleration of block x up the plane. Call this a.

Is the incline frictionless?

You have 2 unknowns T and a which you should be able to solve for with the equations you get.

 

[emma3001]

yes, the incline is frictionless and block y is hanging vertically. if i know that the normal force of y is 31.6N, does that help me in any way?

 

[learningphysics]

yes, the incline is frictionless and block y is hanging vertically. if i know that the normal force of y is 31.6N, does that help me in any way?

If block y is hanging... how is there a normal force?

The way you described the problem... there are 2 forces acting on y... the weight, and tension... write the \Sigma f = ma equation for y.

 

[emma3001]

oops... i guess i meant the gravitational force is 31.6N.

 

[emma3001]

if i only have the weight of block y how am i able to find out FT? After all, isn't FT calculated by saying

Fnet=FT-Fg

do i not have 2 variables here?

 

[learningphysics]

if i only have the weight of block y how am i able to find out FT? After all, isn't FT calculated by saying

Fnet=FT-Fg

do i not have 2 variables here?

yes... I'll call FT, T...

Fnet = T - my*g

now... I'm going to take a as the downward acceleration... taking up as positive and down as negative:

my*(-a) = T - my*g (which has two variables, a and T)

can you come up with an equation for block x?