Block A is 4.5 kg and Block B is 2.25 kg.

Question

Block A is 4.5 kg and Block B is 2.25 kg.

a. determine the mass of block C that must be placed on block A to keep it from sliding if the coefficient of static friction between block A and the table is 0.2

b. If the coefficient of kinematic friction between the surface and block A is 0.15, what must the mass of block C be in order to move across the surface at a constant speed?

c. If block C is lifted, what is the acceleration of the block system if the coefficient of friction between block A and the table is 0.15

Answer 1

Perhaps there is a pulley and block B is hanging ?

call tension in line = T
T = 2.25 (g-a)

Static friction force = .2(4.5+mc)g
so
T - .2 (4.5+mc)g = (4.5+mc) a
so

2.25 (g-a) -.2(4.5+mc)g = (4.5+mc)a

Now in PART A
a = 0
2.25 g -.2(4.5+mc)g = 0
2.25 = .9 + mc
so
mc = 1.35 Kg
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PART B is the same EXCEPT use 0.15 instead of .2
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For Part C, mc = 0 but leave a (and g in the equations

Answer 2

To solve these problems, we will use Newton's second law of motion and the equations related to the force of friction. The force of friction between two surfaces can be calculated using the equation:

Frictional force (Ff) = coefficient of friction (μ) * normal force (Fn)

𝜇 is the coefficient of friction which can be static (μs) or kinetic (μk) depending on the scenario.

a) To determine the mass of block C required to keep block A from sliding, we need to find the force of friction between block A and the table. Since the block is not sliding, the force of friction is in the opposite direction of the applied force and is equal to the maximum force of static friction. The equation is:

Maximum force of static friction = 𝜇s * Fn

Fn is the normal force acting on block A, which is equal to the weight of block A.

Weight of block A = mass of block A * gravitational acceleration (g)

So, the equation becomes:

Maximum force of static friction = 𝜇s * mass of block A * g

Since the maximum force of static friction prevents block A from sliding, it is equal to the force applied by block C. And the force applied by block C can be calculated as:

Force applied by block C = mass of block C * g

Setting these two equations equal, we can solve for the mass of block C:

𝜇s * mass of block A * g = mass of block C * g

Mass of block C = (𝜇s * mass of block A)

b) In this scenario, we want to find the mass of block C required to move block A across the surface at a constant speed. Here, the force of friction is the kinetic friction, which is given by:

Force of kinetic friction = 𝜇k * Fn

Again, Fn is the normal force acting on block A. Since the block is moving at a constant speed, the net force on block A is zero.

Force applied by block C - Force of kinetic friction = 0
mass of block C * g - 𝜇k * mass of block A * g = 0

Solving for the mass of block C:

mass of block C = (𝜇k * mass of block A)

c) If block C is lifted, the friction force no longer acts on block A. In this case, the acceleration of the block system can be calculated using Newton's second law:

Sum of forces = mass * acceleration

The sum of forces acting on block A is the force applied by block C minus the force of friction. The equation becomes:

Force applied by block C - Force of kinetic friction = mass of block A * acceleration

Solving for acceleration:

acceleration = (Force applied by block C - Force of kinetic friction) / mass of block A