Question #74248

A horizontal rod with a mass of 10kg and length 12m hinged to a wall at one end and supported by a cable which make an angle of 30deg with the rod at its other end. Calculate the tension in the cable and force exerted by hinge .Draw the diagram.
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Expert's answer

2018-03-05T09:35:07-0500

Question #74248, Physics / Mechanics | Relativity

A horizontal rod with a mass of 10kg10\mathrm{kg} and length 12m12\mathrm{m} hinged to a wall at one end and supported by a cable which make an angle of 30deg with the rod at its other end. Calculate the tension in the cable and force exerted by hinge. Draw the diagram.

Solution

MA=0\sum M_{A} = 0

Tsin30×(AB)mg×AB2=0;T \sin 3 0 {}^ {\circ} \times (A B) - m g \times \frac {A B}{2} = 0;T=mg2sin30=10×9.812sin30=98.1NT = \frac {m g}{2 \sin 3 0 {}^ {\circ}} = \frac {1 0 \times 9 . 8 1}{2 \sin 3 0 {}^ {\circ}} = 9 8. 1 \mathrm {N}Fx=0;\sum F _ {x} = 0;RAxTcos30=0;R _ {A x} - T \cos 3 0 {}^ {\circ} = 0;RAx=Tcos30=98.1×cos30=85.0NR _ {A x} = T \cos 3 0 {}^ {\circ} = 9 8. 1 \times \cos 3 0 {}^ {\circ} = 8 5. 0 \mathrm {N}Fy=0;\sum F _ {y} = 0;RAy+Tsin30mg=0;R _ {A y} + T \sin 3 0 {}^ {\circ} - m g = 0;RAy=mgTsin30=98.198.1×sin30=49.0NR _ {A y} = m g - T \sin 3 0 {}^ {\circ} = 9 8. 1 - 9 8. 1 \times \sin 3 0 {}^ {\circ} = 4 9. 0 \mathrm {N}RA=RAx2+RAy2=W24tan260+W24=85.02+49.02=98.1NR _ {A} = \sqrt {R _ {A x} ^ {2} + R _ {A y} ^ {2}} = \sqrt {\frac {W ^ {2}}{4 \tan^ {2} 6 0 {}^ {\circ}} + \frac {W ^ {2}}{4}} = \sqrt {8 5 . 0 ^ {2} + 4 9 . 0 ^ {2}} = 9 8. 1 \mathrm {N}


Answer: T=98.1N;RA=98.1NT = 98.1 \, \text{N}; R_A = 98.1 \, \text{N}.

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