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Two blocks sliding!

  • #1

Homework Statement



Two blocks, of weights 3.6 N and 7.5 N, are connected by a massless string and slide down a 30° inclined plane. The coefficient of kinetic friction between the lighter block and the plane is 0.10; that between the heavier block and the plane is 0.20. Assuming that the lighter block leads, find (a) the magnitude of the acceleration of the blocks and (b) the tension in the string.

Homework Equations



Fnet = ma

The Attempt at a Solution



Doing the free body diagram for the heavy weight,

x-axis: Fn(Heavy) - Fg(heavy)sin(theta) = 0 so, Fn(heavy) = Fg(heavy)sin(theta)

y-axis: Fk(heavy) - Ft - Fg(heavy)cos(theta) = -m(heavy)a

For the light weight i got,

x-axis: Fn(light) - Fg(light)sin(theta) = 0 so, Fn(light) = Fg(light)sin(theta)

y-axos: Ft + Fk(light) - Fg(light)cos(theta) = -m(light)a

I equated each block equation to equal to the Ft to get:

M(heavy)a + fk(heavy) - Fg(hanging)cos(theta) = Fg(light)cos(theta) - fk(light) - m(light)a

I set it equal to acceleration but I get the wrong answer.

I also got the mass of each weight by dividing it by g (9.8 m/s^2) Can this be my mistake in the equation?
 
Last edited:

Answers and Replies

  • #2
rl.bhat
Homework Helper
4,433
7
Your last equation is wrong.
The acceleration of the heavy block is given as
[ Fk(heavy) - Ft - Fg(hang)cos(theta)]/M = -a
Similarly write down the acceleration for light block.
 
  • #3
I setted them both equal to the tension and solved for a.
 
  • #4
rl.bhat
Homework Helper
4,433
7
I setted them both equal to the tension and solved for a.
Will you show the calculations, please?
 
  • #5
M(heavy)a + fk(heavy) - Fg(hanging)cos(theta) = Fg(light)cos(theta) - fk(light) - m(light)a

The tension force from the heavy block and the light block.

a = [Fg(light)cos(theta) - Uk(light)Fn(light) - Uk(heavy)Fn(Heavy) + Fg(heavy)cos(theta)] / (m(hang) + m(light)

The answer in the back should be 3.5 m/s^2
 

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