University:
Angelo State University
Course:
ENGR 2301 | Engineering Mechanics - Statics
Academic year:

2024
Views:

500

Pages:

1
Author:

Zoie Hudson

PROBLEM 3.69 A couple M of magnitude 10 lb-ft is applied to the handle of a screwdriver to tighten a screw into a block of wood. Determine the magnitudes of the two smallest horizontal forces that are equivalent to M if they are applied (a) at corners A and D, (b) at corners B and C. (c) anywhere on the block. SOLUTION P D (a) Have or M = Pd hat II =P( 10in .)( 1 ft 12 in. ) 10 lb-P = 12|b| P min =12.001 P A D Min M D P Have (b) d theta*C = sqrt (BE)^ 2 +(EC)^ 2 = sqrt ( 10in .)^ 2 +(6 in.)^ 2 =11.6619 in. M = Pd fl=P( 11.6619in .)( 1\# 12 in. ) 10 lb-P = 10.2899lb or P = 10.29 lb D M P (c) d AC = sqrt (AD)^ 2 +(DC)^ 2 = sqrt (10 in.)^ 2 +(16 in.)^ 2 = 2sqrt(89) * in M=Pd AC ft=P( 2sqrt(89) * in .)( 1 ft 12 in. ) 10 lb-Have P = 6.36lb or P = 6.36