Design of Machinery - SOLUTIONS MANUAL by Robert L. Norton

By Robert L. Norton

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A. When operating them to grab the ice block. b. When clamped to the ice block but before it is picked up (ice grounded). c. When the person is carrying the ice block with the tongs. Solution: See Figure P2-6 and Mathcad file P0225. 1. 1c (Kutzbach's modification) to calculate the mobility. a. In this case there are two links and one full joint and 1 DOF. Number of links L  2 Number of full joints J1  1 Number of half joints J2  0 M  3  ( L  1 )  2  J1  J2 b. When the block is clamped in the tongs another link and two more full joints are added reducing the DOF to zero (the tongs and ice block form a structure).

There are 6 links, 7 full pin joints, and no half-joints. 1c) Number of links L  6 Number of full joints J1  7 Number of half joints J2  0 M  3  ( L  1 )  2  J1  J2 M1 DESIGN OF MACHINERY - 5th Ed. SOLUTION MANUAL 2-50-1 PROBLEM 2-50 Statement: Find the mobility of the mechanism shown in Figure 3-35. Solution: See Figure 3-35 and Mathcad file P0250. 1. 1c to determine the DOF (mobility). There are 6 links, 7 full pin joints, and no half-joints. 1c) Number of links L  6 Number of full joints J1  7 Number of half joints J2  0 M  3  ( L  1 )  2  J1  J2 M1 DESIGN OF MACHINERY - 5th Ed.

The input fourbar consists of links 1, 2, 3, and 4. The alternate output dyad consists of links 5 and 6. The cross-hatched pivot pins at O2, O4 and O6 are attached to the ground link (1). r1  87 r2  49 r3  100 r4  153 B 3 We can analyze this linkage if we replace the slider ( 4) with an infinitely long binary link that is pinned at B to link 3 and pinned to ground (1). Then links 1 and 4 for are both infinitely long. Since these two links are equal in length and, if we say they are finite in length but very long, the rotability of the mechanism will be determined by the relative lengths of 2 and 3.

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