An open vessel of water accelerates up a 30° plane at 3.66 m/s2. What is the angle the water surface makes with the horizontal?

Question

An open vessel of water accelerates up a 30° plane at 3.66 m/s². What is the angle the water surface makes with the horizontal?

Accepted Answer

Referring to Fig. 5-3, the forces acting on each mass dM are weight W vertically downward and force F exerted by the surrounding particles of liquid. This force F is normal to the liquid surface because no frictional component is acting. The resultant force F_x (due to W and F) for each particle of liquid must be up the plane XX at an angle of α = 30° with the horizontal and must cause the common acceleration a_x. Figure 5-3(b) shows this vector relationship. The following equations can now be established:

[latex]F_{x}=\frac{W}{g}a_{x} [/latex] or [latex]\frac{F_{x}}{W}=\frac{a_{x}}{g} [/latex] (1)

F_x sin α = F cos θ - W (2)

F_x cos α = F sin θ from the vector diagram (3)

Multiplying (2) by sin θ and (3) by cos θ and solving simultaneously,

F_x sin α sin θ + W sin θ — F_x cos α cos θ = 0 and [latex]\frac{F_{x}}{W}=\frac{\sin heta }{\cos \alpha \cos heta -\sin \alpha \sin heta } [/latex]

Substituting in (1) and simplifying,

[latex]\frac{a_{x}}{g} =\frac{1}{(\cos \alpha )(\cot heta )-\sin \alpha } [/latex] (4)

from which, since α = 30°,

[latex]\cot heta = an 30^{\circ }+\frac{g}{a_{x}\cos 30^{\circ }}=0.577+\frac{9.81}{3.66 imes 0.866}=3.67 [/latex] and θ = 15°14′ (A)

Note: For a horizontal plane, angle α becomes 0° and equation (4) becomes a/g = tanθ, the equation given for horizontally accelerated motion. For acceleration down the plane, the sign in front of tan 30° becomes minus in equation (A).

Question 5.4: An open vessel of water accelerates up a 30° plane at 3.66 m...

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