This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: wolz (cmw2833) HW #4 Antoniewicz (57420) 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 7 kg 5 kg F F r If F = 29 N and F r = 3 N, what is the magnitude of the force exerted on the block with mass 7 kg by the block with mass 5 kg? Correct answer: 11 . 6667 N. Explanation: m 1 m 2 m 3 F F r Given : vector F = +29 N , vector F r = 3 N , m 1 = 3 kg , m 2 = 7 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 m 2 F 32 F 12 m 3 F r F 23 Let F , 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 F 21 = + F 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 F r = ( m 1 + m 2 + m 3 ) a Solving for a, we have a = F F r m 1 + m 2 + m 3 (4) = 29 N 3 N 3 kg + 7 kg + 5 kg = 1 . 73333 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 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 ( m 1 + m 2 ) F r m 1 + m 2 + m 3 (5) = (5 kg) (29 N) 3 kg + 7 kg + 5 kg + (3 kg + 7 kg) (3 N) 3 kg + 7 kg + 5 kg = 11 . 6667 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 . 73333 m / s 2 ) + 3 N = 11 . 6667 N . 002 (part 1 of 3) 4.0 points wolz (cmw2833) HW #4 Antoniewicz (57420) 2 Two forces, 469 N at 10 and 404 N at 28 are applied to a car in an effort to accelerate it. 3101 kg 4 6 9 N 10 4 4 N 28 What is the magnitude of the resultant of these two forces? Correct answer: 825 . 709 N. Explanation: Given : m = 3101 kg , F 1 = 469 N , 1 = 10 , F 2 = 404 N , and 2 = 28 . Consider the side forces: F 1 ,y = F 1 sin 1 F 2 ,y = F 2 sin 2 F y,net = (469 N) sin10 + (404 N) sin( 28 ) = 108 . 226 N Consider the forward forces: F 1 ,x = F 1 cos 1 F 2 ,x = F 2 cos 2 F x,net = (469 N) cos10 + (404 N) cos( 28 ) = 818 . 586 N Thus the net force is F net = radicalBig (818 . 586 N) 2 + ( 108 . 226 N) 2...
View
Full
Document
 Spring '08
 Turner
 Physics

Click to edit the document details