How Does Friction Affect the Motion of Blocks on Dual Inclined Planes?

[bassmann]

Here's one from my HW:

Two blocks are connected by a massless string on a triangle with a pulley at the top corner. The angle of the left corner is 35 degrees. the angle on the right is 60 degrees. the left block equals 7kg and the right is 9 kg. mu k = 0.25

Determine acceleration of block A and B

Will the tension exceed 70 N?

What is the magnitude and direction of frictional force on A? on B?

 

[Andrew Mason]

Here's one from my HW:

Two blocks are connected by a massless string on a triangle with a pulley at the top corner. The angle of the left corner is 35 degrees. the angle on the right is 60 degrees. the left block equals 7kg and the right is 9 kg. mu k = 0.25

Determine acceleration of block A and B

Will the tension exceed 70 N?

What is the magnitude and direction of frictional force on A? on B?

The frictional force on each block is determined by the component of gravitational force perpendicular to the inclined surface and the coefficient of friction.

So resolve into gravitation force components perpendicular to and along the inclined surfaces. The forces on the blocks are gravity, friction and tension. Tension depends on the acceleration of the other block, which is a function of the friction and gravity forces on it. Let \alpha = 35\degree, \beta = 60\degree angles.

So:
F_{1gr}-F_{1fr}-T=m_1a_1
m_1gsin\alpha - Km_1gcos\alpha - T = m_1_a_1

Similarly:
m_2gsin\beta - Km_2gcos\beta - T = m_2_a_2

The recognition that T is the same for both blocks, enables you to express the equation in terms of both accelerations. Since |a₁|=|a₂| you can figure out the acceleration. Then substitute acceleration back into one of the equations to get T.

AM

 

[wais]

In this dual inclined plane problem, we have two blocks, A and B, connected by a massless string on a triangle with a pulley at the top corner. The left corner has an angle of 35 degrees while the right corner has an angle of 60 degrees. The left block has a mass of 7kg while the right block has a mass of 9kg. The coefficient of kinetic friction is 0.25.

To determine the acceleration of blocks A and B, we can use Newton's second law, which states that the net force acting on an object is equal to its mass multiplied by its acceleration. In this case, the net force acting on both blocks is the tension in the string, and it is the same for both blocks since they are connected by a string. Therefore, we can set up the following equation:

T - μk(mA + mB)g = (mA + mB)a

where T is the tension in the string, μk is the coefficient of kinetic friction, g is the acceleration due to gravity, mA and mB are the masses of blocks A and B respectively, and a is the acceleration of both blocks.

Solving for a, we get:

a = (T - μk(mA + mB)g) / (mA + mB)

Substituting the given values, we get:

a = (T - 0.25(7 + 9)9.8) / (7 + 9)

a = (T - 49) / 16

Next, we can use the fact that the string is inextensible, which means that the acceleration of both blocks must be the same. Therefore, we can set up another equation using the components of the acceleration in the x and y directions:

ax = ay

Since the acceleration of block A is parallel to the incline, we can write:

ax = a sin 35

Similarly, for block B, we can write:

ax = a sin 60

Equating these two equations, we get:

a sin 35 = a sin 60

Solving for a, we get:

a = 0

This means that the acceleration of both blocks is 0, which makes sense since the blocks are not moving in this system.

Moving on to the second part of the problem, we need to determine if the tension in the string will exceed 70N. To do this, we