Some Examples of Static Equilibrium — Question 2

Section 12-4 · Physics pages 1834–1835

Question 2 Cable capacity sets the shelf payload before hinge balance does A horizontal shelf of length 2 m and weight 100 N is hinged to a wall at its left end. A cable from its right end pulls upward and leftward at 30◦above horizontal. A point payload P is at the right end. Cable tension must not exceed 400 N. 100 N P Schematic; labels, not drawing lengths, specify the data. (a) Find the maximum payload and the corresponding cable tension. (b) Find horizontal and vertical hinge reactions at that payload.
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Question 2 - Solution Cable capacity sets the shelf payload before hinge balance does 100 N P (a) Derivation and calculation About the hinge, only the cable vertical component and the weights contribute: 2T sin 30◦−100(1) −2P = 0. Hence, with forces in newtons, T = 100 + 2P. The tension limit gives 100 + 2P ≤400 =⇒ Pmax = 150 N . (b) Result and interpretation At maximum payload, horizontal and vertical balance give Hx −T cos 30◦= 0, Hy + T sin 30◦−100 −P = 0. Thus Hx = 200 √ 3 = 346 N right , Hy = 100 + 150 −400(.5) = 50 N up . Physical check The cable supports 200 N vertically and the hinge supplies the remaining 50 N.

Question 2

Cable capacity sets the shelf payload before hinge balance does

A horizontal shelf of length 2m2\,\mathrm m and weight 100N100\,\mathrm N is hinged to a wall at its left end. A cable from its right end pulls upward and leftward at 30∘30^\circ above horizontal. A point payload PP is at the right end. Cable tension must not exceed 400N400\,\mathrm N.

Diagram shown in the original SVG worksheet.

  1. Find the maximum payload and the corresponding cable tension.

  2. Find horizontal and vertical hinge reactions at that payload.

Question 2 - Solution

Cable capacity sets the shelf payload before hinge balance does

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

About the hinge, only the cable vertical component and the weights contribute: 2Tsin⁡30∘−100(1)−2P=0.2T\sin30^\circ-100(1)-2P=0. Hence, with forces in newtons, T=100+2P.T=100+2P. The tension limit gives 100+2P≤400⇒Pmax=150N.100+2P\le400\quad\Longrightarrow\quad\boxed{P_{\max}=150\ \mathrm N}.

(b) Result and interpretation

At maximum payload, horizontal and vertical balance give Hx−Tcos⁡30∘=0,Hy+Tsin⁡30∘−100−P=0.H_x-T\cos30^\circ=0,\qquad H_y+T\sin30^\circ-100-P=0. Thus Hx=2003=346Nright,\boxed{H_x=200\sqrt3=346\ \mathrm N\ \text{right}}, Hy=100+150−400(.5)=50Nup.\boxed{H_y=100+150-400(.5)=50\ \mathrm N\ \text{up}}.

Physical check

The cable supports 200 N vertically and the hinge supplies the remaining 50 N.