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Home » Engineering Mechanics » Principles of Statics » Resultant of Non-Concurrent Force System

Problem 262 | Resultant of Non-Concurrent Force System

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two-step pulley

Problem 262
Determine completely the resultant of the forces acting on the step pulley shown in Fig. P-262.

Unbalanced forces in two-step pulley

Solution 262

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$R_x = \Sigma F_x$

$R_x = 750 \sin 60^\circ + 250$

$R_x = 899.52 \, \text{ lb to the right}$

$R_y = \Sigma F_y$

$R_y = 750 \cos 60^\circ - 1250$

$R_y = -875 \, \text{ lb}$

$R_y = 875 \, \text{ lb downward}$

$R = \sqrt{{R_x}^2 + {R_y}^2}$

$R = \sqrt{899.52^2 + 875^2}$

$R = 1254.89 \, \text{ lb}$

$\tan \theta_x = \dfrac{R_y}{R_x}$

$\tan \theta_x = \dfrac{875}{899.52}$

$\theta = 44.21^\circ$

$M_{axle} = \Sigma M_{center}$

$M_{axle} = 250(1.25) + 1250(0.5) - 750(1.25)$

$M_{axle} = 0$

Thus, R = 1254.89 lb downward to the right at θ_x = 44.21° and passes through the axle.

‹ Problem 260 - 261 | Resultant of Non-Concurrent Force System up Problem 263 | Resultant of Non-Concurrent Force System ›

Submitted by Romel Verterra on May 11, 2011 - 5:50pm

12091 reads

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