Explanation Assumption: Apply the sign convention for calculating the equations of equilibrium as below. For the horizontal forces equilibrium condition, take the force acting towards right side as positive ( → + ) and the force acting towards left side as negative ( ← − ) . For the vertical forces equilibrium condition, take the upward force as positive ( ↑ + ) and downward force as negative ( ↓ − ) . For moment equilibrium condition, take the clockwise moment as negative and counter clockwise moment as positive. Calculation: Draw the free body diagram of the handle CD as in Figure (1). Refer to Figure (1). Consider the force in the member AC is denoted by A f . The member AC is a two force member. Calculate the distance AC using the relation: A C = 84 2 + 30 2 = 89.1964 mm Consider the horizontal and vertical reaction at D are D x and D y . Consider the sum of the moment about the point D is zero. Then, ∑ M D = 0 − 300 N × ( 126 mm ) − ( A f × 84 mm 89.1964 mm ) × 6 mm + ( A f × 30 mm 89.1964 mm ) × 30 mm = 0 − 37 , 800 N ⋅ mm − ( A f × 84 mm 89.1964 mm ) × 6 mm + ( A f × 30 mm 89.1964 mm ) × 30 mm = 0 − A f × 5.650 mm + A f × 10.09 mm = 37 , 800 N ⋅ mm A f × 4

Vector Mechanics for Engineers: Statics and Dynamics

11th Edition

ISBN: 9780073398242

Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self

Publisher: McGraw-Hill Education

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Chapter 6.4, Problem 6.148P

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Box A has a mass of 15 kilograms and is attached to the 20 kilogram Box B using the cord and pulley system shown. The coefficient of kinetic friction between the boxes and surface is 0.2 and the moment of inertia of the pulley is 0.5 kg * m^ 2. After 2 seconds, how far do the boxes move? A бро B

Box A has a mass of 15 kilograms and is attached to the 20 kilogram Box B using the cord and pulley system shown. The coefficient of kinetic friction between the boxes and surface is 0.2 and the moment of inertia of the pulley is 0.5 kg * m^2. Both boxes are 0.25 m long and 0.25 m high. The cord is attached to the bottom of Box A and the middle of box B. After 2 seconds, how far do the boxes move? A From бро B

Chapter 6 Solutions

Vector Mechanics for Engineers: Statics and Dynamics

Ch. 6.1 - 6.1 through 6.8 Using the method of joints,...Ch. 6.1 - 6.1 through 6.8 Using the method of joints,...Ch. 6.1 - Prob. 6.3P Ch. 6.1 - 6.1 through 6.8 Using the method of joints,...Ch. 6.1 - Prob. 6.5P Ch. 6.1 - Using the method of joints, determine the force in...Ch. 6.1 - 6.1 through 6.8 Using the method of joints,...Ch. 6.1 - Prob. 6.8P Ch. 6.1 - 6.9 and 6.10 Determine the force in each member of...Ch. 6.1 - Prob. 6.10P

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