Question: User: P \theta B 0.4 m 0.5 m 0.4 m G 0.2 m 60° 0.4 m 0.1 m A x y Graph/shape description: A wheeled hand truck with a tilted rectangular crate is shown in the center-right of the image; force P is applied at the top-left handle near point B at angle \theta above the horizontal, with point A at the wheel on the bottom-right and center of mass G inside the crate. position_hint=centerThe hand truck and its contents have a mass of 30 kg with center of gravity at G. (Figure 1) Part A Determine the magnitude of the minimum force required at the grip B needed to lift the load. Express your answer to three significant figures and include the appropriate units. P = 61.3 N Incorrect; Try Again; 5 attempts remaining Part B Determine the direction of the minimum force required at the grip B needed to lift the load. Express your answer in degrees to three significant figures. \theta = 150 °, measured clockwise from the negative x axis. Incorrect; Try Again; 3 attempts remaining Part C Complete previous part(s) graph/figure: A hand truck problem with the load and center of gravity G referenced by Figure 1, positioned top-left.

1. **Problem Statement:** A hand truck with a load of mass $30$ kg has a center of gravity at point $G$. A force $P$ is applied at point $B$ at an angle $\theta$ above the horizontal to lift the load. We need to find: - (A) The minimum magnitude of force $P$ required to lift the load. - (B) The direction $\theta$ of this minimum force. 2. **Given Data:** - Mass $m = 30$ kg - Gravitational acceleration $g = 9.81$ m/s$^2$ - Distances from the figure: $AB = 0.4 + 0.5 = 0.9$ m (horizontal), $AG = 0.4 + 0.2 = 0.6$ m (vertical), and other relevant lengths as per the figure. - Angle between the crate and horizontal is $60^\circ$. 3. **Step 1: Calculate the weight $W$ of the load** $$W = mg = 30 \times 9.81 = 294.3\ \text{N}$$ 4. **Step 2: Set up moment equilibrium about point $A$ (wheel contact point):** To just lift the load, the moment caused by force $P$ about $A$ must balance the moment caused by the weight $W$. Let $P$ act at point $B$ at angle $\theta$ above horizontal. The moment arm for $W$ about $A$ is the horizontal distance from $A$ to $G$, which is $x_G = 0.4 + 0.2 = 0.6$ m. The moment arm for $P$ about $A$ is the vertical and horizontal distances from $A$ to $B$: - Horizontal distance $x_B = 0.4 + 0.5 = 0.9$ m - Vertical distance $y_B = 0.4 + 0.1 = 0.5$ m 5. **Step 3: Write moment equilibrium equation:** Taking counterclockwise moments as positive, $$P \times (y_B \cos \theta + x_B \sin \theta) = W \times x_G$$ 6. **Step 4: Express $P$ in terms of $\theta$:** $$P = \frac{W x_G}{y_B \cos \theta + x_B \sin \theta}$$ 7. **Step 5: Find $\theta$ that minimizes $P$:** The denominator $D = y_B \cos \theta + x_B \sin \theta$ must be maximized to minimize $P$. Using the identity: $$D = \sqrt{x_B^2 + y_B^2} \sin(\theta + \phi)$$ where $$\phi = \arctan\left(\frac{y_B}{x_B}\right) = \arctan\left(\frac{0.5}{0.9}\right) \approx 29.05^\circ$$ Maximum $D$ occurs when $\sin(\theta + \phi) = 1$, so $$\theta + \phi = 90^\circ \Rightarrow \theta = 90^\circ - 29.05^\circ = 60.95^\circ$$ 8. **Step 6: Calculate minimum $P$:** $$D_{max} = \sqrt{0.9^2 + 0.5^2} = \sqrt{0.81 + 0.25} = \sqrt{1.06} = 1.0296$$ $$P_{min} = \frac{294.3 \times 0.6}{1.0296} = \frac{176.58}{1.0296} = 171.5\ \text{N}$$ 9. **Step 7: Convert $\theta$ to the problem's reference frame:** The problem states $\theta$ is measured clockwise from the negative x-axis. Since $\theta$ above is from positive x-axis, the angle from negative x-axis clockwise is: $$\theta_{final} = 180^\circ - 60.95^\circ = 119.05^\circ$$ 10. **Final answers:** - (A) Minimum force magnitude: $$P = 171.5\ \text{N}$$ - (B) Direction of force: $$\theta = 119.1^\circ$$ measured clockwise from the negative x-axis.