Block B weighs 708 N. The coefficient of static friction between block and table is 0.32. Assume that the cord between B and the knot is horizontal. Find the maximum weight of block A for which the system will be stationary.
To find the maximum weight of block A for which the system will be stationary, we need to consider the forces acting on the blocks and use the concept of static equilibrium.
1. Draw a free-body diagram for block B:
- The weight of block B acts downward, represented by the symbol "mg" (mass x acceleration due to gravity).
- The normal force from the table acts upward, perpendicular to the surface.
- The tension in the cord passes through the knot and is horizontal.
2. Identify the forces acting on block B:
- Weight of block B (Wb) = 708 N (given)
- Normal force (N) = equal and opposite to the weight of block B
- Tension in the cord (T) = unknown
3. Write the equations of static equilibrium for block B:
- Sum of forces in the horizontal direction = 0
T = 0 (since Tension is the only force in the horizontal direction)
- Sum of forces in the vertical direction = 0
N - Wb = 0 (Wb = 708 N)
N = 708 N
4. Determine the maximum weight that block A can have without moving block B:
To prevent block B from moving, the maximum frictional force (Ff_max) between block B and the table can be calculated using the equation:
Ff_max = μN
- Coefficient of static friction (μ) = 0.32 (given)
- Normal force (N) = 708 N (calculated in step 3)
Ff_max = 0.32 * 708 N
Ff_max = 226.56 N
Since the tension (T) in the cord is zero (as calculated in step 3) and the maximum frictional force (Ff_max) is less than the weight of block A, the maximum weight of block A for which the system will be stationary is 226.56 N.
To find the maximum weight of block A for which the system will be stationary, we need to consider the forces acting on the system.
Let's start by drawing a free-body diagram for the system:
```
|‾‾‾‾‾‾‾‾| ← Tension force
| Block A |
|_________|
↑ ↑
Frictional │ │ Weight of A
force │ │
_____│__│______
| |
| | ← Weight of B
| Block B |
| |
|_____________|
↑
Normal force
```
The forces acting on block A are its weight and the tension force in the cord. The forces acting on block B are its weight, the tension force in the cord, and the frictional force between block B and the table.
Now let's analyze the forces mathematically:
For block A:
Weight of A = mass of A × acceleration due to gravity
= m_A × g
For block B:
Weight of B = mass of B × acceleration due to gravity
= m_B × g
The tension force in the cord is the same for both blocks since they're connected:
Tension force = T
The frictional force between block B and the table is given by:
Frictional force = coefficient of static friction × normal force
To keep the system stationary, the maximum weight of block A should not exceed the frictional force:
Frictional force ≥ Weight of A
Substituting the values and rearranging the equation, we can solve for the maximum weight of block A:
coefficient of static friction × normal force ≥ m_A × g
We can find the normal force by considering the vertical forces on block B:
Normal force = Weight of B + Weight of A
Now let's calculate the values:
1. Find the normal force:
Normal force = m_B × g + m_A × g
2. Substitute the normal force into the equation for frictional force:
coefficient of static friction × (m_B × g + m_A × g) ≥ m_A × g
3. Simplify the equation:
coefficient of static friction × (m_B + m_A) ≥ m_A
4. Solve for the maximum weight of block A:
m_B ≥ m_A × (1 / coefficient of static friction - 1)
Now, plug in the given values:
m_B ≥ 708 N / (0.32) - 708 N
Solving this equation will give you the maximum weight of block A for which the system will be stationary.