The J-shaped member shown in the figure is supported by a cable DE and a single

Question

The J-shaped member shown in the figure is supported by a cable DE and a single journal bearing with a square shaft at A. Determine the reaction forces and at support A required to keep the system in equilibrium. The cylinder has a mass = 2.90 , and = 7.50 is a vertical force applied to the member at C. The dimensions of the member are = 1.50 , = 6.00 , and = 2.00 . Express your answers numerically in Newtons to three significant figures separated by a comma. Image is at http://www.imagesup.net/?di=213573602319

Explanation / Answer

Tz = Tsin 16.6° … Ty = - Tcos 16.6° sin 63.4°
… Tx = Tcos 16.6° cos 63.4° … balancing of forces gives … along the x-axis …

… Ax = Tx ... along the y-axis .. Ay = Ty … along the z-axis .. Az + Tz – Fb – Fc = 0

… Az = Fb + Fc – Tz … only is torque along the x-axis (due to Fb , Fc , Tz )…

… - 6 ( Fb +Fc ) + 6 Tz = 0 … Tz = Fb + Fc = 5.90 lbs + 2.00 lbs = 7.90 lb … that is

… Az = Fb + Fc – Tz = Fb + Fc – Tz = 0 … using Tz = Tsin 16.6° gives …

… T = Tz / sin 16.6° = 7.90 lb / sin 16.6° = 27.65 lbs … that is …

… Ay = | Ty | = Tcos 16.6° sin 63.4° = ( 27.65 lbs ) cos 16.6° sin 63.4° = 23.7 lbs …

where | Ty | has been taken since the equality refers to the magnitudes on

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