Question

the force $p = 8$ lb is applied to a shift lever as shown below. determine the moment (lb.in) of p about b when $\alpha$ is equal to $17^\circ$. assume the clockwise moment as positive.
your answer:

Explanation:

Step1: Resolve force into components

Horizontal component: $P_x = P\cos\alpha = 8\cos(17^\circ)$
Vertical component: $P_y = P\sin\alpha = 8\sin(17^\circ)$

Step2: Calculate moment from $P_x$

Moment from $P_x$: $M_{P_x} = -P_x \times 22$ (counterclockwise, negative)
$M_{P_x} = -8\cos(17^\circ) \times 22$

Step3: Calculate moment from $P_y$

Moment from $P_y$: $M_{P_y} = P_y \times 8$ (clockwise, positive)
$M_{P_y} = 8\sin(17^\circ) \times 8$

Step4: Sum moments for total moment

$M_B = M_{P_x} + M_{P_y}$
$M_B = -8\times22\cos(17^\circ) + 8\times8\sin(17^\circ)$
Calculate values:
$\cos(17^\circ)\approx0.9563$, $\sin(17^\circ)\approx0.2924$
$M_B \approx -8\times22\times0.9563 + 8\times8\times0.2924$
$M_B \approx -167.4368 + 18.7136$

Answer:

$\approx -148.72$ lb.in