- 2070 Asar · 5 marks
Force P to start motion of bar and block
Short answer
Equilibrium of bar gives wall normal NB = 289.6 N and friction FB = 77.6 N, so pin forces on block are 289.6 N horizontal and 501.6 N vertical. Block ground normal N = 701.6 N, floor friction = 140.3 N. P = 289.6 + 140.3 = 429.9 N, about 430 N to the right.
A uniform bar AB, weighing 424 N, is fastened by a frictionless pin to a block weighing 200 N as shown in figure. At the vertical wall, μ = 0.268 while under the block, μ = 0.20. Determine the force P needed to start motion to the right. [Figure: bar AB inclined at 45° with end B against the vertical wall and end A pinned to the block on the floor; P acts horizontally to the right on the block.]
Reading of the figure: the uniform bar AB (424 N) leans at 45° with end B against the vertical wall on the right (μ = 0.268) and end A pinned to the block (200 N), which is on the floor (μ = 0.20). P acts horizontally to the right on the block. As P pushes the block to the right, B slides up the wall, so wall friction F_B acts downward, and the floor friction under the block acts to the left.
Bar AB (forces: weight 424 N at its mid-point, wall reaction N_B (to the left), wall friction F_B = 0.268 N_B (down), pin force (A_x, A_y) from the block)
Take moments about A (L = length of the bar; B is L cos45° to the right and L sin45° above A):
ΣM_A = 0: N_B (L sin45°) − F_B (L cos45°) − 424 (L/2) cos45° = 0
N_B (1 − 0.268) = 424/2 = 212
N_B = 289.6 N
F_B = 0.268 × 289.6 = 77.6 N
ΣFx = 0: A_x = N_B = 289.6 N (force of block on bar, to the right)
ΣFy = 0: A_y = 424 + F_B = 501.6 N (force of block on bar, upward)
Block (200 N)
The bar pushes the block with 289.6 N to the left and 501.6 N downward.
ΣFy = 0: N = 200 + 501.6 = 701.6 N
Floor friction: F = 0.20 × 701.6 = 140.3 N (to the left)
ΣFx = 0: P = 289.6 + 140.3 = 429.9 N
Answer: P ≈ 430 N.
✓ Answer checked by Claude Opus 5.5 (high effort)
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