Question 5 of 6: Question 5 (paper Question V) — Wedge Blocks (Part B · Dynamics, 20 marks)

Question 5 (paper Question V) — Wedge Blocks (Part B · Dynamics, 20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Block A (wedge, weight 35 N) rests on a frictionless horizontal surface and is pushed by horizontal force P = 45 N; block B (weight 70 N) rests on A's inclined top surface (15° to horizontal) and is confined against a vertical wall (frictionless everywhere, including the A–B interface).

Find. The acceleration of block B.

Approach. A can only accelerate horizontally (it slides on the floor); B can only accelerate vertically (the wall blocks horizontal motion). The frictionless incline contact links the two accelerations kinematically; combine that constraint with each block's own equation of motion.

1. Kinematic constraint. As A advances horizontally by $\Delta s$, the height of A's inclined surface under B's fixed horizontal location rises by $\Delta s\tan\theta$, so $$a_B = a_A\tan\theta$$
2. Equation of motion for A (horizontal; N is the A–B contact normal force, its horizontal component opposes P): $$m_A a_A = P - N\sin\theta, \qquad m_A = \dfrac{35}{9.81} = 3.568\ \text{kg}$$
3. Equation of motion for B (vertical only; the wall supplies whatever horizontal reaction is needed, so it drops out of the vertical equation): $$m_B a_B = N\cos\theta - m_B g, \qquad m_B = \dfrac{70}{9.81} = 7.135\ \text{kg}$$
4. Solve the three equations simultaneously (constraint + 2 equations of motion, 3 unknowns $a_A, a_B, N$): $$N = \boxed{85.20\ \text{N}}, \qquad a_A = \boxed{6.43\ \text{m/s}^2}\ (\text{horizontal}), \qquad a_B = \boxed{1.72\ \text{m/s}^2}\ (\text{vertical, upward})$$