Sand is discharged at A from a conveyor belt and falls onto the top of a stockpile at B. Knowing that the conveyor belt forms an angle α = 25° with the horizontal, determine (a) the speed ν0 of the belt, (b) the radius of curvature of the trajectory described by the sand at Point B.

Question

Sand is discharged at A from a conveyor belt and falls onto the top of a stockpile at B. Knowing that the conveyor belt forms an angle [latex]\alpha =25°[/latex] with the horizontal, determine (a) the speed [latex] u_{ 0 }[/latex] of the belt, (b) the radius of curvature of the trajectory described by the sand at Point B.

Accepted Answer

The motion is projectile motion. Place the origin at Point A. Then [latex]x_{0}=0[/latex] and [latex]y_{0}=0[/latex].

The coordinates of Point B are [latex]x_{B}=30 \mathrm{\ ft}[/latex] and [latex]y_{B}=-18 \mathrm{\ ft}[/latex].

Horizontal motion:

[latex]\begin{aligned} u_{x} & = u_{0} \cos 25^{\circ} & ext{(1)} \x & = u_{0} t \cos 25^{\circ} & ext{(2)}\end{aligned} [/latex]

Vertical motion:

[latex]\begin{aligned} u_{y} & = u_{0} \sin 25^{\circ}-g t & ext{(3)} \y & = u_{0} t \sin 25^{\circ}-\frac{1}{2} g t^{2} & ext{(4)}\end{aligned}[/latex]

At Point B, Eq. (2) gives

[latex] u_{0} t_{B}=\frac{x_{B}}{\cos 25^{\circ}}=\frac{30}{\cos 25^{\circ}}=33.101 \mathrm{\ ft}[/latex]

Substituting into Eq. (4),

[latex]\begin{aligned}-18 & =(33.101)\left(\sin 25^{\circ} ight)-\frac{1}{2}(32.2) t_{B}^{2} \t_{B} & =1.40958 \mathrm{~s}\end{aligned}[/latex]

(a) Speed of the belt.

[latex] u_{0}=\frac{ u_{0} t_{B}}{t_{B}}=\frac{33.101}{1.40958}=23.483[/latex]

[latex] u_{0}=23.4 \mathrm{\ ft} / \mathrm{s}\blacktriangleleft[/latex]

Eqs. (1) and (3) give

[latex]\begin{aligned} u_{x} & =23.483 \cos 25^{\circ}=21.283 \mathrm{\ ft} / \mathrm{s} \ u_{y} & =(23.483) \sin 25^{\circ}-(32.2)(1.40958)=-35.464 \mathrm{~ft} / \mathrm{s} \ an heta \frac{- u_{y}}{ u_{x}} & =1.66632 \quad heta=59.03^{\circ} \ u & =41.36 \mathrm{\ ft} / \mathrm{s}\end{aligned}[/latex]

Components of acceleration.

[latex]\begin{gathered}\mathbf{a}=32.2 \mathrm{\ ft} / \mathrm{s}^{2}\downarrow , \quad a_{t}=32.2 \sin heta \a_{n}=32.2 \cos heta=32.2 \cos 59.03^{\circ}=16.57 \mathrm{\ ft} / \mathrm{s}^{2}\end{gathered}[/latex]

(b) Radius of curvature at B.

[latex]\begin{aligned}& a_{n}=\frac{ u^{2}}{ ho} \& ho=\frac{ u^{2}}{a_{n}}=\frac{(41.36)^{2}}{16.57} & ho=103.2 \mathrm{\ ft}\blacktriangleleft\end{aligned}[/latex]

Question 11.191: Sand is discharged at A from a conveyor belt and falls onto ...

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