Two blocks on ramps, with friction: A marble block of mass m1 = 539.1 kg and a granite block of mass m2 = 134.

Question

Two blocks on ramps, with friction: A marble block of mass m1 = 539.1 kg and a granite block of mass m2 = 134.4 kg are connected to each other via a rope that is directed over a pulley as shown. Both blocks are located on top of inclined planes with angles α = 35.8 degrees and β = 55.2 degrees, as shown. The rope glides over the pulley without friction, but the coefficients of friction of block 1 and the ramp is μ1 = 0.11, and that between block 2 and the ramp is μ2 = 0.25. (For simplicity, let us assume that the static and kinetic friction coefficients are the same in each case.)

What is the acceleration of the marble block? Please note that the positive x-direction is indicated.

Answer 1

Since the diagram is not included, I will assume that the marble is to the right on the ramp set at 35.8 degrees sloping downward to the right , and the granite is to the left on the ramp set at 55.2 degrees sloping downward to the left. You state that the blocks are at the top of the ramps, which is a bit confusing, so I assume that for the solution I present, the blocks are both sliding on their ramps, with positive motion to the right

Using free-body diagrams in a frame of reference each with positive parallel to the specific ramp to the right:

The forces acting on marble while in motion:

friction:
-COS(35.8)*0.11*9.81*539.1= -471.8305451

component of gravity parallel to ramp
sin(35.8)*9.81*539.1= 3093.590196

Tension in rope
-F
Since the block is in motion there is
F=m*a, which is the net
=539.1*a


The forces acting on granite while in motion:

friction:
COS(55.2)*0.25*9.81*134.4=188.1163233

component of gravity parallel to ramp
-sin(55.2)*9.81*134.4= -1082.655671

Tension in rope
F

Since the block is in motion there is
F=m*a, which is the net
=134.4*a


We now have two equations and two unknowns:

a=(2621.76-F)/539.1

a=(F-894.539)/134.4


539.1*F-539.1*894.539
=2621.76*134.4-F*134.4

(539.1+134.4)*F
=539.1*894.539+2621.76*134.4

F=834610.5/673.5

a=2.56 m/s^2
This is parallel to the ramp.

j

Answer 2

As the coef. of static friction is given, it is worth checking to see if the system will move at all .. Friction between m1 and table .. Ff = μ(s).R .. .. (R = perp. force, m1g) Ff = 0.44 x (41kg x 9.80) .. .. .. Ff = 176.80 N When the system is at rest the tension in the string, T = m2.g (weight of m2) T = 15kg x 9.80 .. .. T = 147.0 N The tension is not sufficient to move m1 from rest .. ►T remains at 147.0 N