2 Blocks on an incline?

Question

2 identical blocks tied together w/ a string which passes over a pulley at the crest of inclined planes, one makes an angle theta1 = 26 degrees to the horizontal, the other makes the complementary angle theta2 = 64 degrees.

a) if there is no friction anywhere, with what acceleration do the blocks move?

b) if each block has a mass m=0.6 kg, what is the tension in the string while they are both moving?

c) suppose the coefficient of sliding friction between the blocks and planes is mew= 0.02. With what acceleration do the blocks move?

Answer 1

a)
System equation
F1= mg sin(64)
F2= mg sin(26)

Ft=F1-F2= mg(sin(64) - sin(26))
a=Ft/(m1+m2)= Ft/(2m)
a=mg(sin(64) - sin(26))/(2m)

a=.5g(sin(64) - sin(26))

(b) Tension = ma= 5gm(sin(64) - sin(26))

(c) Now Ft = F1 - f1 - (F2 - f2)

Ft= mg(sin(64) - sin(26)) + umg(cos(24) - cos(26))

where u is the coefficient of friction

Just plug the numbers in

and have fun

Answer 2

? in accordance to capacity conservation regulation: pot capacity of the spring 0.5*ok*x^2 = 0.5*m*v^2 kin contributors capacity of the block, for this reason v = sqrt(ok*x^2/m) = sqrt(4 hundred * 0.22^2 / 2) = 20*0.22/sqrt(2) = 3.111 m/s; ? in accordance to capacity conservation regulation: kin contributors capacity of the block 0.5*m*v^2 = = pot capacity of the spring 0.5*ok*x^2 = = pot capacity of the block mg*h, the place h=L*sin37°; hence 0.5*ok*x^2 = mg*L*sin37°, for this reason ‘how-a techniques’ L=0.5*ok*x^2 / (mg*sin37°) = 0.5*4 hundred*0.22^2 / (2*9.8*sin37°) = 0.821m;