Block B in the figure weighs 739 N. The coefficient of static friction between block and table is 0.34; angle θ is 26°; 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.
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To find the maximum weight of block A for which the system will be stationary, we need to analyze the forces acting on the blocks.
Let's consider block B first. The weight of block B, which is 739 N, can be resolved into two components: one perpendicular to the incline and one parallel to the incline.
The component of the weight perpendicular to the incline is given by Wb⊥ = Wb * cos(θ):
Wb⊥ = 739 N * cos(26°)
Wb⊥ ≈ 666.45 N
The component of the weight parallel to the incline is given by Wb∥ = Wb * sin(θ):
Wb∥ = 739 N * sin(26°)
Wb∥ ≈ 309.18 N
Since the block is in equilibrium, the static frictional force acting on block B must balance the parallel component of the weight. Therefore, the maximum static frictional force can be calculated as:
fstatic(max) = μ * N
where μ is the coefficient of static friction and N is the normal force.
The normal force N can be calculated as:
N = Wb⊥ + Wa
where Wa is the weight of block A.
Now, we can use these values to find the maximum weight of block A.
Since the maximum static frictional force is equal to the parallel component of the weight, we have:
fstatic(max) = Wb∥ = 309.18 N
Substituting the values, we get:
μ * (Wb⊥ + Wa) = 309.18 N
Plugging in the value of the coefficient of static friction (μ = 0.34) and the calculated value of Wb⊥, we have:
0.34 * (666.45 N + Wa) = 309.18 N
Now, let's solve for Wa:
0.34 * 666.45 N + 0.34 * Wa = 309.18 N
226.34 N + 0.34 * Wa = 309.18 N
0.34 * Wa = 309.18 N - 226.34 N
0.34 * Wa = 82.84 N
Wa = 82.84 N / 0.34
Wa ≈ 243.06 N
Therefore, the maximum weight of block A for which the system will be stationary is approximately 243.06 N.
To solve this problem, we need to analyze the forces acting on the system and consider the conditions for static equilibrium. Let's break down the problem step by step:
1. Draw a free-body diagram for both blocks:
+-----------+
| Block A |
+-----------+
<--- w --> (Weight of Block A)
------------------
<--- T ---> | | (Tension in the cord)
| θ |
------------------
<---- N ----> (Normal force on Block B)
+-----------+
| Block B |
+-----------+
<--- W --> (Weight of Block B)
2. Identify the forces acting on each block:
For Block A:
- Weight (w) acting vertically downward.
- Tension (T) acting horizontally towards the right.
For Block B:
- Weight (W) acting vertically downward.
- Normal force (N) acting perpendicular to the inclined surface.
- Frictional force (F) acting parallel to the inclined surface in the opposite direction of Block A.
3. Apply the conditions for static equilibrium:
In the vertical direction:
ΣFy = N - W - w = 0 (No vertical acceleration)
In the horizontal direction:
ΣFx = T - F = 0 (No horizontal acceleration)
4. Calculate the values for N, F, and T:
From the given information, we know that:
- W = 739 N (weight of Block B)
- θ = 26° (angle of the incline)
- μs = 0.34 (coefficient of static friction)
Using the vertical equilibrium equation, we can solve for N:
N - 739 N - w = 0
N = 739 N + w
Using the friction equation, we can solve for F:
F = μs * N
F = 0.34 * (739 N + w) [Substituting the value of N]
From the horizontal equilibrium equation, we can solve for T:
T - F = 0
T = F [Since F is the only horizontal force]
Now we can find the maximum weight of Block A for which the system will be stationary:
Substituting the value of F in terms of w:
T = 0.34 * (739 N + w)
To determine the maximum weight, we need to consider the limiting case where the static friction force (F) is at its maximum value. This limiting case occurs just before the block starts to move:
F_max = μs * N
Substituting the maximum value of F_max and rearranging the equation, we get:
μs * (739 N + w) = μs * N
739 N + w = N
w = N - 739 N
Therefore, the maximum weight of Block A is given by:
Max weight (w) = N - 739
By substituting the value of N from the vertical equilibrium equation, we can determine the maximum weight.