Check — the printed figure carries no coordinate table; pin locations were read in inches from the drawing against its own printed dimension chain (12 in., 12 in., 28 in. horizontally; 9 in., 15 in., 12 in. vertically; 36 in., 16 in. at the roll). This gives A(24,0), B(0,−9), C(12,−24), D(0,−36), E(36,−47.2), F(52,3.6), H(52,−53), with C the frame pivot shared by both arms. Arm CAF (pinned to the frame at C, actuated by cylinder A–B) carries the given 1000 lb at F; arm BCEH (pinned to the frame at C, actuated by cylinder D–E) carries the roll's full 4500 lb weight at its lower support point H. This reconstruction is disclosed here rather than hidden; the METHOD (two rigid bodies, each a 3-equation free body about the shared pin C) is the examinable content.
Given. Two-cylinder roll clamp: arm CAF pinned to the frame at C(12,−24) in., cylinder 1 pinned A(24,0)–B(0,−9), 1000 lb vertical load at F(52,3.6); arm BCEH also pinned to the frame at C, cylinder 2 pinned D(0,−36)–E(36,−47.2), roll weight 4500 lb vertical at H(52,−53). All dimensions in inches, frictionless pins, cylinders are two-force members.
Given data (pin coordinates, in.)
Pin
x
y
A
24
0
B
0
-9
C
12
-24
D
0
-36
E
36
-47.2
F
52
3.6
H
52
-53
Find. a) the force in cylinder 1 (A–B) and cylinder 2 (D–E); b) the pin reaction at C acting on arm BCEH.
Figure 2 — roll-clamp linkage: arm CAF (navy), arm BCEH (teal), cylinders (red), both pinned to the fixed carriage at C.
Approach. Each arm is an independent rigid body pinned to the frame at the common point C, so each can be solved on its own: 3 unknowns (cylinder force + 2 pin-reaction components) against 3 equilibrium equations. Solve arm CAF first (only the given 1000 lb and cylinder 1 act on it), then use cylinder 1's reaction (Newton's third law) plus cylinder 2 and the 4500 lb roll weight to solve arm BCEH.
Arm CAF — set up equilibrium. Cylinder 1 acts at A along the unit vector from A to B, $\hat u_{AB} = (B-A)/|B-A|$; the 1000 lb load acts vertically down at F; the frame reaction is $(C_{x1},C_{y1})$ at C.
$$\hat u_{AB} = (-0.9363,\ -0.3511)$$
Arm CAF — solve. $\sum F_x=0$, $\sum F_y=0$, and $\sum M_C=0$ (3 equations, 3 unknowns $T_1, C_{x1}, C_{y1}$) give
$$T_1 = \boxed{2190.8\ \text{lb}}, \qquad C_{x1}=2051.3\ \text{lb}, \quad C_{y1}=1769.2\ \text{lb}$$
Arm BCEH — transfer the cylinder-1 reaction. By Newton's third law the force cylinder 1 exerts on arm BCEH at B is $-T_1\hat u_{AB}$ (equal, opposite). Cylinder 2 acts at E along $\hat u_{ED}=(D-E)/|D-E|=(-0.9549,\ 0.2971)$; the roll weight 4500 lb acts down at H; frame reaction $(C_{x2},C_{y2})$ at C.
Arm BCEH — solve. $\sum F_x=0$, $\sum F_y=0$, and $\sum M_C=0$ give
$$T_2 = \boxed{14\,644.1\ \text{lb}}, \qquad C_{x2}=-16\,034.3\ \text{lb}, \quad C_{y2}=8081.1\ \text{lb}$$
Substituting all forces back into both arms' equilibrium equations closes to zero — the solution is self-consistent.
Combine for the C reaction on arm BCEH.
$$C_{BCEH} = \sqrt{C_{x2}^2+C_{y2}^2} = \boxed{17\,955.6\ \text{lb}}, \qquad \theta=\arctan\!\left(\dfrac{C_{y2}}{C_{x2}}\right)=153.3^\circ \text{ from +x}$$
