EXAMPLE 9–5 | Forces on a beam and supports. A uniform 1500-kg beam, 20.0 m long, supports a 15,000-kg printing press 5.0 m from the right support column (Fig. 9–8). Calculate the force on each of the vertical support columns. APPROACH We analyze the forces on the beam (the force the beam exerts on each column is equal and opposite to the force exerted by the column on the beam). We label these forces F, and Fg in Fig. 9–8. The weight of the beam itself acts at its center of gravity, 10.0 m from either end. We choose a con- venient axis for writing the torque equation: the point of application of FA (labeled P), so F, will not enter the equation (its lever arm will be zero) and we will have an equation in only one unknown, FR. CG (1500 kg)ỹ 10.0 m -5.0 m-5.0 m- (15,000 kg)ỹ SOLUTION The torque equation, E7 = 0, with the counterclockwise direction as positive, gives FIGURE 9-8 A 1500-kg beam supports a 15,000-kg machine. Example 9–5. Er = -(10.0 m)(1500 kg)g (15.0 m)(15,000 kg)g + (20.0 m)F3 = 0. Solving for FB, we find Fg = (12,000 kg)g = 118,000 N. To find FA, we use EF, = 0, with +y upward: = FA - (1500 kg)g – (15,000 kg)g + Fg EFy = 0. Putting in Fg = (12,000 kg)g, we find that FA (4500 kg)g 44,100 N.
EXAMPLE 9–5 | Forces on a beam and supports. A uniform 1500-kg beam, 20.0 m long, supports a 15,000-kg printing press 5.0 m from the right support column (Fig. 9–8). Calculate the force on each of the vertical support columns. APPROACH We analyze the forces on the beam (the force the beam exerts on each column is equal and opposite to the force exerted by the column on the beam). We label these forces F, and Fg in Fig. 9–8. The weight of the beam itself acts at its center of gravity, 10.0 m from either end. We choose a con- venient axis for writing the torque equation: the point of application of FA (labeled P), so F, will not enter the equation (its lever arm will be zero) and we will have an equation in only one unknown, FR. CG (1500 kg)ỹ 10.0 m -5.0 m-5.0 m- (15,000 kg)ỹ SOLUTION The torque equation, E7 = 0, with the counterclockwise direction as positive, gives FIGURE 9-8 A 1500-kg beam supports a 15,000-kg machine. Example 9–5. Er = -(10.0 m)(1500 kg)g (15.0 m)(15,000 kg)g + (20.0 m)F3 = 0. Solving for FB, we find Fg = (12,000 kg)g = 118,000 N. To find FA, we use EF, = 0, with +y upward: = FA - (1500 kg)g – (15,000 kg)g + Fg EFy = 0. Putting in Fg = (12,000 kg)g, we find that FA (4500 kg)g 44,100 N.
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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Question
If the left vertical support column in Example 9–5 is made of steel, what is its cross-sectional area? Assume that a safety factor of 3 was used in its design to avoid fracture
Transcribed Image Text:EXAMPLE 9–5 | Forces on a beam and supports. A uniform 1500-kg beam,
20.0 m long, supports a 15,000-kg printing press 5.0 m from the right support
column (Fig. 9–8). Calculate the force on each of the vertical support columns.
APPROACH We analyze the forces on the beam (the force the beam exerts on
each column is equal and opposite to the force exerted by the column on the
beam). We label these forces F, and Fg in Fig. 9–8. The weight of the beam
itself acts at its center of gravity, 10.0 m from either end. We choose a con-
venient axis for writing the torque equation: the point of application of FA
(labeled P), so F, will not enter the equation (its lever arm will be zero) and we
will have an equation in only one unknown, FR.
CG
(1500 kg)ỹ
10.0 m -5.0 m-5.0 m-
(15,000 kg)ỹ
SOLUTION The torque equation, E7 = 0, with the counterclockwise direction
as positive, gives
FIGURE 9-8 A 1500-kg beam supports
a 15,000-kg machine. Example 9–5.
Er = -(10.0 m)(1500 kg)g
(15.0 m)(15,000 kg)g + (20.0 m)F3 = 0.
Solving for FB, we find Fg = (12,000 kg)g = 118,000 N. To find FA, we use
EF, = 0, with +y upward:
= FA - (1500 kg)g – (15,000 kg)g + Fg
EFy
= 0.
Putting in Fg = (12,000 kg)g, we find that FA
(4500 kg)g
44,100 N.
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