Explanation Given Information: The unstretched length of spring is 0.3 m. The angle θ is 30 ° . Assumptions: Consider the state of member as tension where the force is pulling the member and as compression where the force is pushing the member. Consider the force indicating right side as positive and left side as negative in horizontal components of forces. Consider the force indicating upside is taken as positive and downside as negative in vertical components of forces. Consider clockwise moment as negative and anti-clock wise moment as positive wherever applicable. Apply the Equation of equilibrium wherever applicable. The acceleration due to gravity ( g ) is 9.81 m/sec 2 . Convert the mass of bar in to weight of bar using the relation. W = m g Here, the mass is m . The weight of each bar is ( m × 9.81 ) N . Member BC : Show the free body diagram of the member BC as in Figure (1). Using Figure (1), Spring force: Determine the elongation of the spring, x using the equation. x = 2 d − Unstretched length (I) Determine the spring force using the equation. F s p = k x (II) Here, the stiffness is k and the compression is x . Moment about point C : Determine the horizontal and vertical reaction at point B by taking moment about point C. ∑ M C = 0 B y ( 2 cos 30 ° ) − B x ( 2 sin 30 ° ) − m × 9.81 ( cos 30 ° ) = 0 2 B y cos 30 ° − 2 B x sin 30 ° = 8.5 m (III) Conclusion: Substitute 2 sin 30 ° for d and 0.3 m for unstretched length in Equation (I). x = 2 ( 2 sin 30 ° ) − 0 .3 = 1.7 m Substitute 150 N/m for k and 1.7 m for x in Equation (II). F s p = 150 × 1

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Free Body Diagram

The system of forces acting on a body tends to either rotate it or make the body undergo motion. In general, when an external agent exerts a force on a body, the body undergoes translational motion and rotational motion. The body has six degrees of freed…

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Chapter 6.6, Problem 111P

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The mass m of each uniform bar for equilibrium

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Chapter 6.6, Problem 110P

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Chapter 6 Solutions

ENGINEERING MECHANICS

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The mass m of each uniform bar for equilibrium

Solution Summary: The author explains how the mass m of each uniform bar for equilibrium is underset_26kg

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