Part A — Minimum distance x (bracket-on-pipe friction)
Given. Pipe diameter 3 in.; bracket sleeve (clamp) height h = 6 in.; coefficient of static friction μs = 0.25 between the pipe and the bracket sleeve; bracket weight neglected.
Find. The minimum horizontal distance x (from the pipe centreline to the load W) at which the bracket will not slip down the pipe.
Figure 3A — bracket clamped to the pipe; normal forces develop at the top and bottom of the 6 in. sleeve.
Approach. At the minimum x the bracket is on the verge of slipping down the pipe, so friction at both the top and bottom contact points is fully mobilized ($f=\mu_s N$) and acts upward. The two contacts push the pipe apart horizontally with equal and opposite normal forces N (statics of the sleeve in the x-direction).
Horizontal equilibrium of the sleeve. The pipe bears on the sleeve at the top (one side) and bottom (opposite side) with equal normal forces:
$$\sum F_x = 0:\quad N_{top} = N_{bottom} = N$$
Vertical equilibrium — verge of slipping. Both friction forces act upward and are fully mobilized:
$$\sum F_y = 0:\quad \mu_s N + \mu_s N - W = 0 \Rightarrow N = \dfrac{W}{2\mu_s}$$
Moment equilibrium about the bottom contact. The top normal force N acts horizontally at a moment arm h above the bottom contact, and W acts at a horizontal distance x from the pipe centreline:
$$\sum M_{bottom} = 0:\quad W\,x = N\,h$$
Combine. Substituting $N = W/(2\mu_s)$:
$$x_{min} = \dfrac{h}{2\mu_s} = \dfrac{6}{2(0.25)} = \boxed{12.00\ \text{in.}}$$
Note x is independent of W — the same minimum distance applies for any load.
Part A result
Quantity
Value
Minimum distance x
12.00 in.
Part B — Centre of gravity of the machine element
Given. A machine element composed of a horizontal base block (4.5 in. × 2 in. × 0.5 in.) with two 1 in.-diameter through-holes, and a vertical block (0.5 in. × 2.5 in. × 2 in.) with a semicylindrical cutout (radius 1 in.) at its outer end; origin at the rear-bottom-left corner.
Given data (composite-body volumes and centroids, in.3, in.)
Part
Volume
x̄
ȳ
z̄
Base block (+)
4.500
2.25
1.00
0.25
Vertical block (+)
2.500
4.25
3.25
1.00
Hole 1 (−)
0.393
1.00
1.00
0.25
Hole 2 (−)
0.393
3.00
1.00
0.25
Semi-cyl. cutout (−)
0.785
4.25
4.076
1.00
Find. The coordinates $(\bar X,\bar Y,\bar Z)$ of the centroid (centre of gravity, uniform material) of the composite machine element.
Figure 3B — composite machine element (isometric sketch): base plate + end flange, two through-holes, one semicylindrical cutout.
Approach. Treat the element as a base block plus a vertical block, minus the two cylindrical holes and the semicylindrical cutout. Each simple solid's volume and centroid are computed directly (holes: $V=\pi r^2 t$; half-cylinder: $V=\tfrac12\pi r^2 t$, with its centroid offset $4r/(3\pi)$ from the flat face), then combined by the composite-body (weighted-average) formula.
Half-cylinder centroid offset. For the semicylindrical cutout (radius 1 in., flat face at y=4.5):
$$\bar y_{semi} = 4.5-\dfrac{4r}{3\pi} = 4.5-\dfrac{4(1)}{3\pi} = \boxed{4.076\ \text{in.}}$$
Total (signed) volume.
$$V = 4.500+2.500-0.393-0.393-0.785 = \boxed{5.429\ \text{in.}^3}$$
First moments and centroid. Summing $V_i\bar x_i$, $V_i\bar y_i$, $V_i\bar z_i$ (holes and cutout negative) and dividing by V:
$$\bar X = \dfrac{\sum V_i\bar x_i}{V} = \boxed{2.918\ \text{in.}}$$
$$\bar Y = \dfrac{\sum V_i\bar y_i}{V} = \boxed{1.591\ \text{in.}}$$
$$\bar Z = \dfrac{\sum V_i\bar z_i}{V} = \boxed{0.487\ \text{in.}}$$
