Explanation Show the free body diagram of the container as in Figure (1). Here, is the weight of the container is w , normal force is N, the angle of inclination of the ground surface with respect to horizontal is θ , a small cube of length dx , height dy , and width dz , and the acceleration of the container is a . Calculate the acceleration of the container, a . F = m a w sin ϕ = ( w g ) a sin ϕ = a g a = g sin ϕ Show the components of acceleration, a x and a y as in Figure (2). Here, the component of acceleration acting along the horizontal axis ( x -axis) is a x and the component of acceleration acting along the vertical axis ( y -axis) is a y . Calculate the components of acceleration, a x and a y by resolving the acceleration acting along sloping direction with respect to the horizontal and vertical axis. a x = a cos ϕ a y = a sin ϕ Substitute a = − ( g sin ϕ ) in the above equations. a x = − ( g sin ϕ ) cos ϕ a y = − ( g sin ϕ ) sin ϕ Assume a small cube of length dx , height dy , and width dz . Show the horizontal and the vertical force acting on the cube as in Figure (3). Refer the Figure (3). Apply the principle of equilibrium of force along the x direction to the cube. → + ∑ F x = m a x p x d y d z − ( p x + ∂ p x ∂ x d x ) d y d z = ( w g ) a x p x d y d z − ( p x + d p x ) d y d z = ( γ V g ) a x ( p x − p x − d p x ) d y d z = ( γ d x d y d z g ) a x ( p x − p x − d p x ) d y d z d y d z = γ d x g a x − d p x = γ d x g a x d p x = − γ d x g a x Here, the pressure acting along the x -direction is p x , the volume is V , and the unit weight of the fluid is γ . Refer the Figure (3). Apply the principle of equilibrium of force along the y direction to the cube. + ↑ ∑ F y = m a y p y d x d z − ( p y + ∂ p y ∂ y d y ) d x d z − w = ( w g ) a y p y d x d z − ( p y + d p y ) d x d z − γ V = ( γ V g ) a y p y d x d z − ( p y + d p y ) d x d z − γ d x d y d z = ( γ d x d y d z g ) a y ( p y − p y − d p y − γ d y ) d x d z = ( γ d x d y d z g ) a y − d p y − γ d y = γ d y g a y − d p y = γ d y + γ d y g a y d p y = − γ d y ( 1 + a y g ) Here, the pressure acting along the y -direction is p y

2nd Edition

ISBN: 9780134649290

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 2, Problem 139P

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Show that the slope of the surface of the liquid (θ) , during the motion is equal to ϕ.

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Chapter 2, Problem 138P

Chapter 2, Problem 140P

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Show that the slope of the surface of the liquid ( θ ) , during the motion is equal to ϕ .

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Show that the slope of the surface of the liquid (θ) , during the motion is equal to ϕ.

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Chapter 2 Solutions