Kinetic Friction coefficient.

[Fire Slayer]

Homework Statement

Blocks 1 and 2 of masses m1 and m2, respectively, are connected by a light string. These blocks are further connected to a block of mass M by another light string that passes over a pulley of negligible mass and friction. Blocks 1 and 2 move with a constant velocity v down the inclined plane, which makes an angle $$\theta$$ with the horizontal. The kinetic frictional force on block 1 is f and that on block 2 is 2f.

determine the coefficient of kinetic friction between the inclined plane and block 1.

and

Determine the value of the suspend mass M that allows the two blocks to move with constant velocity down the plane

Homework Equations

(sum of forces) = (mass)(acceleration)
(kinetic friction force)=(kinetic friction coefficient)(normal force)

The Attempt at a Solution

I set the sum of the forces equal to zero for both the vertical and the horizontal components, but I don't think they came out right.

 

[cristo]

Why don't you post what you got, and someone can check your work for you.

 

[m2003]

I would approach this problem by first considering the definition of kinetic friction coefficient. Kinetic friction coefficient is a measure of the resistance to motion between two surfaces in contact when they are already in motion. It is represented by the symbol μk and is defined as the ratio of the kinetic friction force (f) to the normal force (N) between the two surfaces: μk = f/N.

In this problem, we are given that the kinetic friction force on block 1 is f and that on block 2 is 2f. We can use this information to set up the following equations:

f = μk(N1) and 2f = μk(N2)

Where N1 and N2 are the normal forces acting on block 1 and block 2, respectively.

Next, we can use the fact that the blocks are moving with constant velocity to set up the following equation:

m1g(sinθ) - f = m1a

Where m1 is the mass of block 1, g is the acceleration due to gravity, and a is the acceleration of the blocks down the inclined plane. Similarly, for block 2 we have:

m2g(sinθ) - 2f = m2a

We can solve these equations for a and set them equal to each other, since the blocks are moving with the same velocity:

m1g(sinθ) - μk(N1) = m2g(sinθ) - 2μk(N2)

We can then solve for μk:

μk = (m1g(sinθ) - m2g(sinθ)) / (N1 - 2N2)

Finally, we can use the fact that the blocks are connected by a light string to set up an equation for the tension in the string, T:

T = m1a = m2a

Substituting in our expression for a from the previous equation, we get:

T = (m1g(sinθ) - μk(N1)) / m1 = (m2g(sinθ) - 2μk(N2)) / m2

We can then solve for the value of M that allows the two blocks to move with constant velocity down the inclined plane by setting the two expressions for T equal to each other and solving for M:

(m1g(sinθ) - μk(N1)) / m1 = (m2g(sinθ) -