# Lecture 7 ENGR-1100 Introduction to Engineering Analysis.

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Lecture 7 ENGR-1100 Introduction to Engineering Analysis

Lecture Outline Equilibrium of a particle - three dimensional problems.

Equilibrium of a particle – three dimensions R= R x + R y + R z = = R x i + R y j + R z k =  F x i +  F y j +  F k k The equation can only be satisfied if: R x = R x i=  F x i=0 R y = R y j=  F y j=0 R z = R z j=  F z j=0

Example – P3-22 Struts AB and AC of Fig. P3-22 can transmit only axial tensile compressive forces. Determine the forces in struts AB and AC and the tension in cable AD when force F=1250 N.

Solution 1) Free body diagram – body #1 T AD = (-6/(3 2 +6 2 ) 1/2 j + 3/(3 2 +6 2 ) 1/2 k) T AD =(-0.89j + 0.477k ) T AD T AD F AB F AC F z y x F AC = (-3/(3 2 +6 2 +2 2 ) 1/2 i -6/(3 2 +6 2 +2 2 ) 1/2 j -2/(3 2 +6 2 +2 2 ) 1/2 k )F AC = (-0.43i - 0.86j - 0.29k ) F AC F AB = (2/(2 2 +6 2 +2 2 ) 1/2 i -6/(2 2 +6 2 +2 2 ) 1/2 j -2/(2 2 +6 2 +2 2 ) 1/2 k ) F AB =( 0.3i - 0.9j - 0.3k) ) F AB F = (-1250 k) N

 F x =0 -0.43F AC +0.3F AB =0  F y =0 -0.89T AD -0.86F AC -0.9F AB =0  F z =0 0.477T AD -0.29F AC -0.3F AB =1250 lb T AD F AB F AC F z y x

0 -0.43 0. 3 0 -0.89 -0.86 -0.9 0 0.48 -0.29 -0.3 1250 A= Using Maple: T AD =1606.5 N; F AC =-665.1 N; F AB =-953.2 N;

Class Assignment: Exercise set 3-21 please submit to TA at the end of the lecture The block shown in fig. P3-21 weighs 500 lb. Determine the tensions in cable AB, AC, and AD. Solution: T AB =267.4 lb; T AC =142.5 lb; T AD =164 lb;

T AB = (6/(6 2 +12 2 ) 1/2 j + 12/(6 2 +12 2 ) 1/2 k) T AB =(0.447j + 0.89k ) T AB T AC = (4/(4 2 +3 2 +12 2 ) 1/2 i -3/(4 2 +3 2 +12 2 ) 1/2 j -12/(4 2 +3 2 +12 2 ) 1/2 k )T AC = (0.31i – 0.23j + 0.92k ) T AC T AB T AC T AD z y x W=500 lb T AD = (-4/(4 2 +8 2 +12 2 ) 1/2 i-8/(4 2 +8 2 +12 2 ) 1/2 j+12/(4 2 +8 2 +12 2 ) 1/2 k) T AC = (-0.27i - 0.53j + 0.8k ) T AC Solution

T AB = (0.447j + 0.89k ) T AB T AC = (0.31i – 0.23j + 0.92k ) T AC T AD = (-0.27i - 0.53j + 0.8k ) T AD W = (-500 k) lb  F x =0: 0T AB +0.31T AC -0.27T AD =0  F y =0: 0.447T AB -0.23T AC -0.53T AD =0  F z =0: 0.89T AB +0.92T AC +0.8T AD =500

0 -0.31 -0.27 0 0.45 -0.23 -0.53 0 0.89 0.92 0.8 500 A= Using Maple: T AB =267.4 lb; T AC =142.5 lb; T AD =164 lb;