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Unformatted text preview: rose (kmr2427) – HW #4 – Antoniewicz – (57380) 1 This printout should have 14 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The horizontal surface on which the objects slide is frictionless. The acceleration of gravity is 9 . 8 m / s 2 . 3 kg 8 kg 5 kg F ℓ F r If F ℓ = 23 N and F r = 3 N, what is the magnitude of the force exerted on the block with mass 8 kg by the block with mass 5 kg? Correct answer: 9 . 25 N. Explanation: m 1 m 2 m 3 F ℓ F r Given : vector F ℓ = +23 N ˆ ı, vector F r = − 3 N ˆ ı , m 1 = 3 kg , m 2 = 8 kg , m 3 = 5 kg , and g = 9 . 8 m / s 2 . Note: F is acting on the combined mass of the three blocks, resulting in a common acceleration after accounting for friction. m 1 F 21 F ℓ m 2 F 32 F 12 m 3 F r F 23 Let F ℓ , F r , F 32 represent the force exerted on the system from the right, from the left, and the force exerted on m 2 by m 3 , respec tively. Note: vector F 21 = − vector F 12 and vector F 32 = − vector F 23 , and bardbl vector F 21 bardbl = bardbl vector F 12 bardbl and bardbl vector F 32 bardbl = bardbl vector F 23 bardbl , where bardbl vector F bardbl ≡ F is the magnitude of vector F . The sum of the forces acting on each block separately are m 1 a = + F ℓ − F 21 = + F ℓ − F 12 (1) m 2 a = + F 12 − F 32 = + F 12 − F 23 (2) m 3 a = + F 23 − F r (3) To find the acceleration we can treat the three masses as a single object or add the forces acting on each component of the sys tem, Eqs. 1, 2, and 3. F ℓ − F r = ( m 1 + m 2 + m 3 ) a Solving for a, we have a = F ℓ − F r m 1 + m 2 + m 3 (4) = 23 N − 3 N 3 kg + 8 kg + 5 kg = 1 . 25 m / s 2 . We can solve for F 23 using Eq. 3 and sub stituting a from Eq. 4. The result is F 23 = m 3 a + F r (3) = m 3 ( F ℓ − F r ) m 1 + m 2 + m 3 + F r ( m 1 + m 2 + m 3 ) m 1 + m 2 + m 3 = m 3 F ℓ − ( m 1 + m 2 ) F r m 1 + m 2 + m 3 (5) = (5 kg) (23 N) 3 kg + 8 kg + 5 kg + (3 kg + 8 kg) (3 N) 3 kg + 8 kg + 5 kg = 9 . 25 N . Alternative Solution: Using Eq. 3 and the numerical result of Eq. 4, we have F 23 = m 3 a + F r (3) = (5 kg) (1 . 25 m / s 2 ) + 3 N = 9 . 25 N . 002 (part 1 of 3) 4.0 points rose (kmr2427) – HW #4 – Antoniewicz – (57380) 2 Two forces, 350 N at 12 ◦ and 292 N at 32 ◦ are applied to a car in an effort to accelerate it. 3311 kg 3 5 N 12 ◦ 2 9 2 N 32 ◦ What is the magnitude of the resultant of these two forces? Correct answer: 595 . 648 N. Explanation: Given : m = 3311 kg , F 1 = 350 N , θ 1 = 12 ◦ , F 2 = 292 N , and θ 2 = − 32 ◦ . Consider the side forces: F 1 ,y = F 1 sin θ 1 F 2 ,y = F 2 sin θ 2 F y,net = (350 N) sin12 ◦ + (292 N) sin( − 32 ◦ ) = − 81 . 9673 N Consider the forward forces: F 1 ,x = F 1 cos θ 1 F 2 ,x = F 2 cos θ 2 F x,net = (350 N) cos12 ◦ + (292 N) cos( − 32 ◦ ) = 589 . 982 N Thus the net force is F net = radicalBig (589 . 982 N) 2 + ( − 81 . 9673 N) 2 = 595 . 648 N...
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This note was uploaded on 03/01/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Physics

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