This problem involves finding the minimum required force $P$ to lift the rock block using two wedges (A and B) and rollers. The system is simplified by setting up equilibrium equations for the critical state (impending motion).

1. Identify Parameters and Forces

• Mass of Rock Block: $M = 100 \text{ kg}$
• Weight of Rock Block: $W = M g = 100 \text{ kg} \cdot g$
• Wedge Angle: $\alpha = 10^\circ$
• Coefficient of Static Friction ($\mu_s$): $0.3$ (between A and C, and B and D)
• Friction between rollers and wedges: Ignored (smooth rollers $\implies$ only normal forces at interface $AB$).

The required force $P$ is requested in $\text{kg}$ (mass equivalent of the force $P = M_p g$). We will calculate $P$ in Newtons and then convert the magnitude back to kilograms by dividing by $g$ (or simply ignoring $g$ if we work entirely in mass/force equivalent units).

2. Analyze the Free Body Diagrams (FBDs)

A. FBD of Rock Block (Assumed to be in equilibrium with Wedge B)

The rock block exerts weight $W$ downwards, resisted by the normal force $N_{\text{Rock}}$ from the vertical wall D, and the force $F_{\text{top}}$ from wedge B.

Since the rock block is lifted vertically, $W$ is supported by the vertical component of the force $F_{\text{top}}$ from wedge B. Assuming the rock is rigid and the force is transmitted vertically, the normal force on the top of wedge B ($N_{B, \text{top}}$) is equal to $W$.

$$N_{B, \text{top}} = W = 100 g \text{ N}$$

B. FBD of Wedge B (Top Wedge)

Wedge B is acted upon by:

• $N_1$: Normal force from the rock/wall D (vertical wall). Since the rock is constrained, $N_{\text{Rock}}$ is the horizontal reaction from the wall. We assume the rock is heavy and only transmits vertical force $W$ to $B$.
• $N_{B, \text{bottom}}$: Normal force from rollers/Wedge A, perpendicular to the $10^\circ$ incline.
• $F_{B, D}$: Friction force from vertical wall D. Since B moves right, $F_{B, D}$ acts downwards. $F_{B, D} = \mu_s N_{B, D}$.

Wait, the geometry shows the rock rests on B, and the vertical wall D is fixed.

Let's assume the rock block rests on top of B, and B rests against the vertical wall D. * Vertical forces on B: $W_{\text{rock}} = 100 g$ (down) and $N_{B, \text{bottom}} \cos(10^\circ)$ (up). * The friction $\mu_s$ is between B and D. Since B moves right, $N_{B, D}$ is the horizontal force on D, and $F_{B, D}$ is vertical (downwards). $N_{B, D} = N_{B, \text{bottom}} \sin(10^\circ)$.

This structure is solved by combining the two wedges A and B into a single composite system supporting the rock block $W$ and resisting the force $P$.

3. Combined System Analysis (Wedges A and B)

Since the rollers are smooth (frictionless), the force transmitted between A and B ($F_{AB}$) is purely normal to the surface (at $10^\circ$).

We analyze the two external friction surfaces: $A$ on $C$ ($\mu_s=0.3$) and $B$ on $D$ ($\mu_s=0.3$).

Since the wedges are assumed weightless, the only external vertical load is $W = 100 g$. This entire load is transmitted to the ground $C$ and wall $D$.

Let $R_{AB}$ be the normal force between A and B (rollers). $R_{AB}$ acts perpendicular to the $10^\circ$ incline.

A. FBD of Wedge B (Lifting the Rock): Wedge B is in equilibrium (or impending motion). $W$ acts downwards on B. $N_D$ acts horizontally left from the wall D. * $N_{AB}$ (Normal force from rollers/A): Acts up and left, $10^\circ$ from vertical. * $N_{B, D}$ (Normal force from wall D): Acts horizontally left. * $F_{B, D}$ (Friction on D): Acts downwards, $F_{B, D} = 0.3 N_{B, D}$. * $W = 100 g$. This analysis is complicated. We use the standard principle for friction wedges, where the friction angle $\phi_s = \arctan(\mu_s)$.

$$\phi_s = \arctan(0.3) \approx 16.699^\circ$$

Since the rollers are smooth, we only consider friction at surfaces $\text{A-C}$ and $\text{B-D}$.

A. FBD of Wedge A (Force P applied)

* $P$ (Horizontal, right). * $N_{AC}$ (Normal force from ground C): Vertical up. * $F_{AC}$ (Friction from C): Horizontal left. $F_{AC} = 0.3 N_{AC}$. * $R_{AB}$: Force from B (rollers), acting normal to incline, up and left, $10^\circ$ from vertical.

B. FBD of Wedge B (Lifting the Rock W)

* $W$ (Vertical down). * $N_{B, D}$ (Normal force from wall D): Horizontal left. * $F_{B, D}$ (Friction from D): Vertical down (since B moves right, D resists movement). $F_{B, D} = 0.3 N_{B, D}$. * $R_{AB}$: Force from A, acting normal to incline, down and right, $10^\circ$ from vertical. **Wait, the system is symmetric. The total force $W$ is supported by the vertical components of $F_{AB}$ and the reactions at the top.** Let $R_{AB}$ be the resultant force acting on the wedge B from the rollers (perpendicular to the $10^\circ$ slope). **Simplifying assumption:** Since the rollers are smooth and the two wedges have identical $10^\circ$ slopes, the system acts as a compound wedge where $W$ is resisted by $P$. The friction surfaces are the horizontal ground $C$ and the vertical wall $D$. This is equivalent to a single wedge system where the effective wedge angle is $2\alpha = 20^\circ$. The vertical force $W$ is resisted by $P$. The minimum force $P$ required to lift $W$ using a wedge system with angle $\alpha$ and friction angle $\phi_s$ at the external surfaces is complex. **Standard Formula (Single Wedge against Fixed Wall, $P$ driving the wedge):** $$P = W \tan(\alpha + \phi_s)$$ *Here, $\alpha = 10^\circ$ and $\phi_s = 16.70^\circ$.* $$P_{\text{lifting}} = W \tan(10^\circ + 16.70^\circ) = W \tan(26.70^\circ)$$ $$P_{\text{lifting}} = W \cdot 0.5028$$ $$P_{\text{lifting}} \approx 0.5028 W$$ **Applying to the compound system (Two wedges, A and B):** The external friction is on surfaces C (horizontal) and D (vertical). * FBD B (Top): $W$ down. $N_{AB}$ up-left. $N_D$ left. $F_D$ down. * FBD A (Bottom): $P$ right. $N_{AB}$ down-right. $N_C$ up. $F_C$ left. The effective angle for lifting $W$ by pushing $A$ is $2\alpha = 20^\circ$. If we treat the system as one wedge with an effective angle $2\alpha = 20^\circ$, pushing against a fixed structure: $$P = W \tan(2\alpha + \phi_s)$$ (Incorrect, the friction angle must be applied only where the surfaces rub.) **Let's use the external forces and geometry:** The total lifting force $W$ must be overcome. **FBD of Rock + Wedge B (System $W+B$):** $$\sum F_y = 0: N_{AB} \cos(10^\circ) - W - F_{B, D} = 0$$ $$\sum F_x = 0: N_{B, D} - N_{AB} \sin(10^\circ) = 0$$ $$N_{B, D} = N_{AB} \sin(10^\circ)$$ $$F_{B, D} = 0.3 N_{B, D} = 0.3 N_{AB} \sin(10^\circ)$$ $$N_{AB} \cos(10^\circ) - 100 g - 0.3 N_{AB} \sin(10^\circ) = 0$$ $$N_{AB} [\cos(10^\circ) - 0.3 \sin(10^\circ)] = 100 g$$ $$N_{AB} [0.9848 - 0.3(0.1736)] = 100 g$$ $$N_{AB} [0.9848 - 0.05208] = 100 g$$ $$N_{AB} [0.93272] = 100 g$$ $$N_{AB} \approx 107.215 g$$ **FBD of Wedge A:** $$\sum F_y = 0: N_{AC} - N_{AB} \cos(10^\circ) = 0 \implies N_{AC} = N_{AB} \cos(10^\circ)$$ $$N_{AC} \approx 107.215 g \cdot 0.9848 \approx 105.54 g$$ $$F_{AC} = 0.3 N_{AC} \approx 31.66 g$$ $$\sum F_x = 0: P - F_{AC} - N_{AB} \sin(10^\circ) = 0$$ $$P = F_{AC} + N_{AB} \sin(10^\circ)$$ $$P = 0.3 N_{AC} + N_{AB} \sin(10^\circ)$$ $$P = 31.66 g + 107.215 g \cdot 0.1736$$ $$P = 31.66 g + 18.61 g$$ $$P \approx 50.27 g \text{ N}$$ **Conversion to kg (Mass equivalent of Force P):** $$P_{\text{kg}} = \frac{P}{g} \approx 50.27 \text{ kg}$$ **Rounding off to one decimal place:** $$P_{\text{kg}} = 50.3 \text{ kg}$$ This result falls within the constraint range [50, 51].