1.6 ft 1 Question #4 ( The 250-lb pipe is being towed behind a truck, as shown at left. If 0= 25° and the coefficient of friction of the pipe with the ground is 4 = 0.25, determine: a) the tension in the cable b) the acceleration of the truck

1.6 ft 1 Question #4 ( The 250-lb pipe is being towed behind a truck, as shown at left. If 0= 25° and the coefficient of friction of the pipe with the ground is 4 = 0.25, determine: a) the tension in the cable b) the acceleration of the truck

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Using the component method, determine the resultant force and the equilibrant force of each of the following sets of concurrent forces. 1. F₁ = 380 N @ 50⁰ N of E F2 = 650 N @ 22⁰ E of S F3 = 720 N West 3. FA = 20 lb S А FB = 24 lb NE Fc = 16 lb West FD = 25 lb @ 32º E of S A FOLLOW THE FORMAT BELOW Example 2: F₁ = 200 d West F₂ = 300 d @ 35⁰ W of N F3 = 250 d SE F₁ = 450 d @ 80⁰ S of W Σ Fx -200 (West) 300(sin 35⁰) =-172 (West) 250(cos 45°) = +177 (East) 450(cos 80°) = -78.1 (West) 2. F₁ = 332 d @ 35⁰ W of N F2 = 420 d @ 75⁰ E of N F3 = 108 d SW -273 (West) Fy 0 300(cos 35°) =+246 (North) 45⁰ = -177 (South) 450(sin 80°) =-443 (South) -374 (South) Sketch F₁ = 200 d F₂=300 d FAY! F3-250 d 45⁰ F t 80⁰ 35⁰1 F2y F3v F4=150 d ANSWER NO. 3 ONLY
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1. A 20 N crate is held at rest on an incline by a rope that is parallel to the incline as shown on the figure to the right. The incline is 30° above the horizontal. The coefficient of static friction, µs, between the crate and the incline is 0.35. a) Draw two free body diagrams for the forces acting on the crate. In the first, indicate the forces pointing along the ramp and in the second, indicate the forces pointing perpendicular to the ramp b) Rank all forces acting on the crate from largest to smallest. c) Write down two separate equations for Newton's second law with static friction: one for forces pointing along the ramp and one for forces pointing perpendicular to the ramp. Do some forces appear in only one of the Newton's law equations? For forces that have some component in both, make sure you break down those force vectors correctly in the equation.
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Using the component method, determine the resultant force and the equilibrant force of each of the following sets of concurrent forces. 1. F₁ = 380 N @ 50⁰ N of E F2 = 650 N @ 22⁰ E of S F3 = 720 N West 3. FA = 20 lb S А FB = 24 lb NE Fc = 16 lb West FD = 25 lb @ 32º E of S A FOLLOW THE FORMAT BELOW Example 2: F₁ = 200 d West F₂ = 300 d @ 35⁰ W of N F3 = 250 d SE F₁ = 450 d @ 80⁰ S of W Σ Fx -200 (West) 300(sin 35⁰) =-172 (West) 250(cos 45°) = +177 (East) 450(cos 80°) = -78.1 (West) 2. F₁ = 332 d @ 35⁰ W of N F2 = 420 d @ 75⁰ E of N F3 = 108 d SW -273 (West) Fy 0 300(cos 35°) =+246 (North) 45⁰ = -177 (South) 450(sin 80°) =-443 (South) -374 (South) Sketch F₁ = 200 d F₂=300 d FAY! F3-250 d 45⁰ F t 80⁰ 35⁰1 F2y F3v F4=150 d ANSWER NO. 2 ONLY
Please draw the free diagram clear and also indicate if we need to add force
Please include illustrations too. Thank you
Using the component method, determine the resultant force and the equilibrant force of each of the following sets of concurrent forces. 1. F₁ = 380 N @ 50⁰ N of E F2 = 650 N @ 22⁰ E of S F3 = 720 N West 3. FA = 20 lb S А FB = 24 lb NE Fc = 16 lb West FD = 25 lb @ 32º E of S A FOLLOW THE FORMAT BELOW Example 2: F₁ = 200 d West F₂ = 300 d @ 35⁰ W of N F3 = 250 d SE F₁ = 450 d @ 80⁰ S of W Σ Fx -200 (West) 300(sin 35⁰) =-172 (West) 250(cos 45°) = +177 (East) 450(cos 80°) = -78.1 (West) 2. F₁ = 332 d @ 35⁰ W of N F2 = 420 d @ 75⁰ E of N F3 = 108 d SW -273 (West) Fy 0 300(cos 35°) =+246 (North) 45⁰ = -177 (South) 450(sin 80°) =-443 (South) -374 (South) Sketch F₁ = 200 d F₂=300 d FAY! F3-250 d 45⁰ F t 80⁰ 35⁰1 F2y F3v F4=150 d ANSWER NO. 2 ONLY
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