A block of mass M resting on a 21.5° slope is shown. The block has coefficients of friction μs=0.605 and μk=0.344 with the surface. It is connected via a massless string over a massless, frictionless pulley to a hanging block of mass 2.57 kg. What is the minimum mass M1 that will stick and not slip?
Apply T=T and use uk an us but u must have uk biger than the force
To find the minimum mass M1 that will stick and not slip, we need to do the following steps:
Step 1: Determine the forces acting on the block of mass M on the inclined plane.
- The weight of the block can be resolved into two components: one parallel to the slope (mg*sin(θ)) and one perpendicular to the slope (mg*cos(θ)).
- The friction force acting on the block can be calculated as fs = μs * N, where N is the normal force.
Step 2: Find the normal force.
- The normal force can be calculated as N = mg*cos(θ).
Step 3: Calculate the maximum static friction force.
- The maximum static friction force is fs_max = μs * N.
Step 4: Calculate the force acting down the slope due to the hanging block.
- The force acting down the slope due to the hanging block is given by F = M1 * g.
Step 5: Determine the maximum force of static friction required for the block not to slip.
- The maximum force of static friction required for the block not to slip is given by fs_max = F.
Step 6: Calculate the minimum mass M1 that will stick and not slip.
- Simplify the equation by substituting the values obtained from the previous steps:
μs * mg*cos(θ) = M1 * g.
Step 7: Solve for M1.
- Divide both sides of the equation by g:
μs * m*cos(θ) = M1.
Substituting the given values, we have:
μs = 0.605
θ = 21.5°
m = 2.57 kg
Solving for M1:
M1 = μs * m*cos(θ)
M1 = 0.605 * 2.57 kg * cos(21.5°)
M1 = 1.289 kg
Therefore, the minimum mass M1 that will stick and not slip is approximately 1.289 kg.
To determine the minimum mass M₁ that will stick and not slip, we need to consider the forces acting on the block.
Let's break down the forces acting on the block:
1. Weight (W): This force is acting vertically downward and is given by W = M₁ * g, where g is the acceleration due to gravity (approximately 9.8 m/s²).
2. Normal force (N): This force is acting perpendicular to the surface of the slope. It can be calculated as N = M₁ * g * cos(θ), where θ is the angle of the slope (21.5° in this case).
3. Frictional force (F_friction): There are two types of friction involved:
a. Static friction (F_friction,static): This force opposes the tendency of the block to initiate motion. It can be calculated as F_friction,static = μs * N, where μs is the coefficient of static friction.
b. Kinetic friction (F_friction,kinetic): This force opposes the motion of the block once it starts sliding. It can be calculated as F_friction,kinetic = μk * N, where μk is the coefficient of kinetic friction.
Now, let's analyze the forces acting on the hanging block:
1. Weight (W'): This force is acting vertically downward and is given by W' = M₂ * g, where M₂ is the mass of the hanging block (2.57 kg).
Taking into account that the masses on either side of the pulley are connected by a string, the tension in the string on both sides of the pulley will be the same (assuming an ideal, massless, and frictionless pulley).
Now we can set up equations to find the minimum mass M₁ that will stick and not slip.
Equation for static friction:
F_friction,static = F_tension,static
μs * N = T₁
μs * M₁ * g * cos(θ) = M₂ * g
By substituting the given values and solving for M₁, we can find the minimum mass:
M₁ = M₂ / (μs * cos(θ))
M₁ = 2.57 kg / (0.605 * cos(21.5°))
Calculating the above expression will give us the value for the minimum mass M₁.