The 15-lb rod BC connects a disk centered at A to crank CD. Knowing that the disk is made to rotate at the constant speed of 180 rpm, determine for the position shown the vertical components of the forces exerted on rod BC by the pins at B and C.

Question

Accepted Answer

We first determine the acceleration of Point B of disk.

[latex]{\boldsymbol{\omega}}=180\mathrm{\ rpm}=18.85\,\mathrm{{rad/s}}[/latex]

Since ω = constant

[latex]a_{B}=(a_{B})_{n}=r\omega^{2}[/latex]

[latex]={\left\lgroup{\frac{8}{12}}{\mathrm{\ ft}}\right\rgroup}(18.85\,{\mathrm{rad/s}})^{2}[/latex]

[latex]{\bf a}_{B}=236.9\;\mathrm{ \ ft}/s^{2}\ ⦫60^{\circ}[/latex]

Since rod BC is in translation.

[latex]\overline{{{\bf a}}}=\overline{{{\bf a}}}_{B}=236.9\;\mathrm{ft}/s^{2}\;⦫60^{\circ}[/latex]

Vertical components of forces at B and C.

[latex]+\circlearrowleft \Sigma M_{B}=\Sigma(M_{B})_{\mathrm{eff}}\colon\ C_{y}(30\mathrm{~in.})-W(15\ \mathrm{in.})=-m\overline{{{a}}}\sin60^{\circ}(15\ \mathrm{in.})[/latex]

Since [latex]\quad W=15 \mathrm{\ lb}\qquad\mathrm{and}\quad m\overline{{{a}}}=\frac{15 \mathrm{\ lb}}{32.2}(236.9)\!\!=110.36 \mathrm{\ lb}[/latex]

[latex]30C_{y}-(15)(15)=-110.36\sin60^{\circ}(15)=-95.57(15)[/latex]

[latex]2C_{y}=-95.57+15=80.57,\;\;\;C_{y}=-40.285\,\mathrm{\ lb}[/latex]

[latex]\mathbf{C}_{y}=40.3 \mathrm{\ lb}\downarrow \blacktriangleleft[/latex]

[latex]+\uparrow\mathbf{\Sigma}F_{y}=\Sigma(F_{y})_{\mathrm{eff}}\colon\qquad\ B_{y}-W+C_{y}=-m\overline{{{a}}}\sin60^{\circ}[/latex]

[latex]B_{y}=W-C_{y}-{\frac{15}{32.2}}(236.9)\sin60^{\circ}=15+40.285-95.57[/latex]

[latex]B_{y}=-40.285\,\mathrm{lb}[/latex]

[latex]\mathbf{B}_{\mathrm{y}}=40.3\,\mathrm{lb}\downarrow\blacktriangleleft[/latex]

Question 16.18: The 15-lb rod BC connects a disk centered at A to crank CD. ...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions