From the given figure below, 500kN is applied to push the wedge out from under the 600kN block (block A). The angle of friction is 25o for all contact surfaces. Determine the Reaction Force of Block B in contact with the plane
To solve this problem, we need to draw a free body diagram of block A and block B.
Free body diagram of block A:
Where:
- m_a = mass of block A
- g = acceleration due to gravity
- F = 500 kN (applied force)
- f = friction force
- R_A = reaction force of the plane on block A
Since the wedge is pushed out from under block A, there is no normal force acting on it. Therefore, the y-component of the reaction force of block A is equal to the weight of block A.
Free body diagram of block B:
Where:
- m_b = mass of block B
- R_B = reaction force of the plane on block B
Since block B is in contact with the plane, friction force is acting on it.
Now, we can solve for the unknowns using the equations from the free body diagrams:
For block A:
- y-component of R_A = m_a * g = 600 kN (given)
- x-component of R_A = 0 (no horizontal force acting on block A)
- z-component of R_A = F - f - m_a * g = 500 kN - f - 600 kN = -100 kN + f
For block B:
- y-component of R_B = m_b * g = R_A = 600 kN (since the two blocks are in contact and not sliding)
- x-component of R_B = 0 (no horizontal force acting on block B)
- z-component of R_B = f
To solve for f, we need to use the equation for friction force:
f = μ * R_A = 0.25 * 600 kN = 150 kN
Substituting in the values for R_A and f in the equations for block A, we get:
- z-component of R_A = -100 kN + f = -100 kN + 150 kN = 50 kN
Therefore, the reaction force of block B in contact with the plane is 600 kN in the y-direction.