the system is in static equilibrium. The coefficient of static friction between the block and the table's surface is = 1/3. The mass of the block that is hanging down is n1 Kg. The tangent of the angle shown in the figure is tan theta = n2. What is the minimum value of mass, n3 Kg, that the block on the table can have such that the system is in equilibrium?
To solve this problem, we need to analyze the forces acting on the block. Let's go step-by-step:
Step 1: Identify the forces acting on the block:
- Weight of the block hanging down: W1 = n1 * g, where g is the acceleration due to gravity.
- Normal force exerted by the table on the block: N.
- Friction force between the block and the table: F_friction = coefficient of static friction * N.
Step 2: Determine the sum of the forces in the horizontal direction:
In equilibrium, the sum of forces in the horizontal direction is zero. Since there are no other horizontal forces acting on the block, the friction force must be equal in magnitude and opposite in direction to the component of the weight projected horizontally. Therefore, we have:
F_friction = n1 * g * sin(theta) (1)
Step 3: Determine the sum of the forces in the vertical direction:
In equilibrium, the sum of forces in the vertical direction is zero. Therefore, we have:
N - n1 * g * cos(theta) = 0 (2)
Step 4: Find the minimum value of mass, n3, for equilibrium:
Since we want to find the minimum value of mass that the block on the table can have such that the system is in equilibrium, we need to find the value of n3 at the point where the block is just about to start sliding. This occurs when the friction force is at its maximum, which is given by the maximum static friction, F_friction_max = coefficient of static friction * N.
Step 5: Substitute F_friction_max into Equation (1):
By substituting F_friction_max into Equation (1) and rearranging the equation, we have:
coefficient of static friction * N = n1 * g * sin(theta)
coefficient of static friction * (n3 * g * cos(theta)) = n1 * g * sin(theta)
n3 = (n1 * sin(theta)) / (coefficient of static friction * cos(theta))
Therefore, the minimum value of mass, n3, that the block on the table can have such that the system is in equilibrium is given by:
n3 = (n1 * sin(theta)) / (coefficient of static friction * cos(theta))
n3 = (n1 * n2) / (1/3 * cos(theta))
n3 = 3n1n2 / (cos(theta))
To find the minimum value of the mass n3 that the block on the table can have such that the system is in equilibrium, we need to consider the forces acting on the system.
First, let's analyze the forces acting on the block hanging down:
1. The weight force (mg1), where g is the acceleration due to gravity and m is the mass of the block hanging down.
2. The tension force in the string (T), which is acting upwards.
Now, let's analyze the forces acting on the block on the table:
3. The weight force (mg3), where g is the acceleration due to gravity and m is the mass of the block on the table.
4. The normal force (N), which is acting upwards from the table's surface.
5. The static friction force (fs), which opposes the tendency of the block on the table to move.
Since the system is in static equilibrium, the sum of the forces in both the x and y directions must add up to zero.
In the vertical direction (y-direction), we have:
mg1 - T - mg3 + N = 0
But since the system is in equilibrium, N must be equal to mg3. Therefore, we have:
mg1 - T - mg3 + mg3 = 0
mg1 - T = 0
mg1 = T
Next, in the horizontal direction (x-direction), we have:
fs = T
Now, let's look at the given information:
- The coefficient of static friction between the block and the table's surface is 1/3.
- The tangent of the angle (tan θ) shown in the figure is n2.
From the given information, we can say that the static friction force (fs) is given by:
fs = μsN
where μs is the coefficient of static friction and N is the normal force.
Since N = mg3, we have:
fs = μsmg3
But we also know that fs = T, so we can write:
T = μsmg3
From earlier, we know that mg1 = T, so we can write:
mg1 = μsmg3
Now, substituting n1 for m1 and n3 for m3, we have:
n1g = μsn3g
n1 = μsn3
Finally, we can solve for the minimum value of mass n3 by simplifying the equation:
n3 = n1 / (μs * tan θ)
Therefore, the minimum value of mass n3 that the block on the table can have such that the system is in equilibrium is n1 / (μs * tan θ) kg.