Question

the crowbar is subjected to a vertical force of p = 27 lb at the grip, whereas it takes a force of f = 165 lb at the claw to pull the nail out. the crowbar contacts the board at point a. (figure 1)

Explanation:

Step1: Take moments about point A.

Sum of moments $\sum M_A=0$. The moment of a force $M = Fd\sin\theta$, where $F$ is the force, $d$ is the perpendicular - distance from the point of rotation to the line of action of the force, and $\theta$ is the angle between the force and the position vector from the point of rotation to the point of application of the force.
The moment due to force $P$ about point $A$ is $M_P=P\times(14 + 1.5)\sin(90^{\circ}-20^{\circ})$. The moment due to force $F$ about point $A$ is $M_F=-F\times3\sin60^{\circ}$.

Step2: Set up the moment - equilibrium equation.

$P\times(14 + 1.5)\sin70^{\circ}-F\times3\sin60^{\circ}=0$.
We know that $P = 27$ lb and we want to check if the given $F = 165$ lb satisfies the equation.
Left - hand side of the equation:
\[

$$\begin{align*} LHS&=27\times(14 + 1.5)\sin70^{\circ}-165\times3\sin60^{\circ}\\ &=27\times15.5\times0.9397-165\times3\times0.8660\\ &=27\times14.5654 - 165\times2.598\\ &=393.2658-428.67\\ &\approx - 35.4 \end{align*}$$

\]
Since the left - hand side is not equal to 0, the system is not in equilibrium with the given values.

If we want to find the correct value of $F$ for equilibrium:
\[

$$\begin{align*} F&=\frac{P\times(14 + 1.5)\sin70^{\circ}}{3\sin60^{\circ}}\\ &=\frac{27\times15.5\times0.9397}{3\times0.8660}\\ &=\frac{27\times14.5654}{2.598}\\ &=\frac{393.2658}{2.598}\\ &\approx151.37\text{ lb} \end{align*}$$

\]

Answer:

The value of $F$ for equilibrium is approximately $151.37$ lb.