A man stacks a small block of wood on top of a larger block and then moves both by pushing the
bottom block. The large block has a mass of 25 kg, and the smaller block has a mass of 15 kg. The
coefficient of static friction between the two blocks is 0.45. How fast can the man accelerate the blocks
without having the smaller block slide off the top?
To determine the maximum acceleration without the smaller block sliding off the top, we need to calculate the maximum force of static friction between the two blocks.
The formula for static friction is:
Frictional force = coefficient of static friction * Normal force
The normal force is equal to the weight of the smaller block, which is its mass multiplied by the acceleration due to gravity (9.8 m/s²).
Normal force = mass of smaller block * acceleration due to gravity
Normal force = 15 kg * 9.8 m/s²
Normal force = 147 N
The maximum force of static friction can be calculated as:
Maximum frictional force = coefficient of static friction * Normal force
Maximum frictional force = 0.45 * 147 N
Maximum frictional force = 66.15 N
Since the maximum frictional force between the blocks is equal to the force applied to accelerate the blocks, we can use Newton's second law to calculate the maximum acceleration.
Force = mass * acceleration
66.15 N = (25 kg + 15 kg) * acceleration
66.15 N = 40 kg * acceleration
Acceleration = 66.15 N / 40 kg
Acceleration = 1.65375 m/s²
Therefore, the man can accelerate the blocks at a maximum rate of 1.65375 m/s² without having the smaller block slide off the top.
To solve this problem, we will consider the forces acting on the smaller block and the larger block separately.
Let's denote the acceleration of the blocks as a, and let's assume the direction of acceleration is to the right.
For the smaller block:
The only force acting on the smaller block is the static friction force (Fs) between the two blocks. The maximum value of static friction force can be calculated using the equation Fs ≤ μsN, where μs is the coefficient of static friction and N is the normal force between the two blocks. In this case, the normal force is equal to the weight of the smaller block, which is N = mg = 15 kg * 9.8 m/s^2 = 147 N. Plugging in these values, we have Fs ≤ 0.45 * 147 N = 66.15 N.
For the larger block:
The only force acting on the larger block is the force of gravity (weight). The weight of the larger block is W = mg = 25 kg * 9.8 m/s^2 = 245 N.
Since the smaller block can slide off the top if the static friction force is less than the force of gravity, we can set up the inequality Fs ≥ W, which becomes 66.15 N ≥ 245 N.
Therefore, the maximum acceleration the man can apply without the smaller block sliding off is given by Newton's second law: Fs = ma, which becomes 66.15 N = (15 kg + 25 kg) * a. Solving for acceleration, we have a = 66.15 N / 40 kg = 1.65 m/s^2.
Therefore, the man can accelerate the blocks up to a speed of 1.65 m/s^2 without having the smaller block slide off the top.
To determine the maximum acceleration at which the smaller block will not slide off the top of the larger block, we need to consider the forces acting on the system.
1. First, let's calculate the force of static friction (F_static) between the two blocks. The formula for static friction is:
F_static = coefficient of static friction * normal force
2. The normal force (N) can be found by considering the weight of the smaller block and the force applied by the larger block. The weight (W) of an object is given by:
W = mass * gravity
where the mass is the mass of the smaller block and the gravity is the acceleration due to gravity (approximately 9.8 m/s²).
3. The force applied by the larger block (F_applied) can be calculated using Newton's second law of motion:
F_applied = mass * acceleration
4. Finally, to prevent the smaller block from sliding off, the force of static friction should be equal to or greater than the force applied:
F_static ≥ F_applied
Now, let's put it all together to find the maximum acceleration.
Step 1: Calculate the normal force (N)
N = W_smaller_block + W_larger_block
W_smaller_block = mass_smaller_block * gravity
W_larger_block = mass_larger_block * gravity
Step 2: Calculate the force of static friction (F_static)
F_static = coefficient_of_static_friction * N
Step 3: Substitute F_static and F_applied into the inequality
F_static ≥ F_applied
coefficient_of_static_friction * N ≥ mass * acceleration
Step 4: Solve for the maximum acceleration (acceleration_max)
acceleration_max = (coefficient_of_static_friction * N) / mass
Now we can substitute the given values into the equation to find the maximum acceleration.
mass_smaller_block = 15 kg
mass_larger_block = 25 kg
coefficient_of_static_friction = 0.45
gravity = 9.8 m/s²
Step 1: Calculate the normal force (N)
W_smaller_block = mass_smaller_block * gravity = 15 kg * 9.8 m/s²
W_larger_block = mass_larger_block * gravity = 25 kg * 9.8 m/s²
N = W_smaller_block + W_larger_block
Step 2: Calculate the force of static friction (F_static)
F_static = coefficient_of_static_friction * N
Step 3: Substitute F_static and F_applied into the inequality
coefficient_of_static_friction * N ≥ mass * acceleration
Step 4: Solve for the maximum acceleration (acceleration_max)
acceleration_max = (coefficient_of_static_friction * N) / mass
By plugging in the values, you can calculate the maximum acceleration.