--- title: "Force P to start motion of bar and block" subject: "Applied Mechanics" chapter: 5 key: "semester_1/applied_mechanics:bar-ab-pin-block-force-p-start-motion" source: https://www.bctengineer.com/q/semester_1/applied_mechanics/bar-ab-pin-block-force-p-start-motion site: BCT Engineering --- # Force P to start motion of bar and block Applied Mechanics · Chapter 5: Friction - Asked in: 2070 Asar, 5 marks - Video: [Determine horizontal force P necessary to cause motion. | Chapter 6: Friction | Engineers Academy](https://www.youtube.com/watch?v=N5XoofjiV5I) — Engineers Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/applied_mechanics/bar-ab-pin-block-force-p-start-motion - Whole chapter: https://www.bctengineer.com/md/semester_1/applied_mechanics/chapter-5 ## Question 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.] ## 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. ## Answer **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): ```text 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. ```text 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.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. As P pushes the block right, which way does wall friction F_B act on end B of the bar? A. Downward, since B slides up the wall B. Upward, since B slides up the wall C. Towards the right D. Zero, as the wall is vertical Answer: A. Friction opposes sliding. B moves up the wall, so friction acts downward. (Key point: Direction of wall friction) 2. Taking moments about A for the bar, what is the result for N_B? A. N_B = 424 N B. N_B = 289.6 N C. N_B = 212 N D. N_B = 501.6 N Answer: B. N_B(1 - 0.268) = 424/2 = 212, so N_B = 289.6 N. (Key point: Moment equation on bar) 3. What is the normal reaction N under the block? A. 200 N B. 424 N C. 701.6 N D. 289.6 N Answer: C. The bar pushes down with A_y = 424 + F_B = 501.6 N, so N = 200 + 501.6 = 701.6 N. (Key point: Normal reaction under block) 4. How is P found once the floor friction is known? A. P = horizontal push of bar (289.6 N) + floor friction (140.3 N) = 429.9 N B. P = 0.20 x 200 = 40 N C. P = 289.6 - 140.3 = 149.3 N D. P = 0.268 x 701.6 = 188 N Answer: A. On the block, P balances the bar's 289.6 N leftward push and the leftward floor friction 0.20 x 701.6 = 140.3 N, giving about 430 N. 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