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Unformatted text preview: STATIC EQUILIBRIUM E XERCISES Section 12.1 Conditions for Equilibrium 14. (a) ˆ ˆ ˆ ˆ ˆ (2 2 2 3 )N 0. i F i j i j j ∑ = + + = v (b) ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ( ) 2 (2 2 ) ( ) ( 2 3 ) 7 ) ] N m i i i j i i j i j j ∑ = [ × + +  ×  + ( + × ⋅ = r τ ˆ (4 3 7) N m 0. k + ⋅ = 15. I NTERPRET We have been told that the choice of pivot point does not matter if the sum of forces is zero. Here we will show that this is true for two different pivot points. Three forces are acting on an object, which is in equilibrium. D EVELOP The three forces are 1 ˆ ˆ 2 2 N F i j = + r at point ( , ) (2 m, 0 m), x y = 2 ˆ ˆ 2 3 N F i j =  r at ( 1 m, 0 m), and 3 ˆ 1 N F j = r at ( 7 m, 1 m). We find the torques due to these three forces around points (3 m, 2 m) and ( 7 m, 1 m), using   . r F τ = × r r r To find the value of r r for a point other than the origin, we take the vector difference between the point where the force is applied and the point used as the pivot: applied pivot ( ). r r r = r r r E VALUATE For point (3 m, 2 m), the torque due to 1 F r is 1 applied pivot 1 1 ˆ ˆ ˆ ˆ ( ) ((2 3) m (0 2) m) (2 2 ) N ˆ ˆ ˆ ˆ ˆ 1 2 0 Nm ( 2 ( 4)) Nm 2 Nm 2 2 r r F i j i j i j k k k τ τ = × = + × + =  =    = r r r r r Similarly, 2 ˆ 8 Nm k τ = r and 3 ˆ 10 Nm. k τ =  r The sum of these three is ˆ (2 8 10) Nm 0. total k τ = + = r Around point (–7 m, 1 m), the torques are 1 ˆ 20 Nm, k τ = r 2 ˆ 20 Nm, k τ =  r and 3 0. τ = r The sum of these three is also 0. total τ = r A SSESS Note that the torque due to force 3 around the second pivot is zero, since the force acts on the pivot. 16. The conditions for static equilibrium, under the action of three forces, can be written as: 3 1 2 ( ) F F F v v v = + and 3 3 1 1 2 2 ( ). r F r F r F × =  × + × v v v v v v (a) In this case, 1 1 2 ˆ ˆ ˆ , (2 m) , , F F j r j F Fi = = = v v v v v and 2 ˆ (1 m) . r j = v Thus, 3 ˆ ˆ ( ), F F i j =  + v which is a force of magnitude 2 , 45 F ° down into the third quadrant ( 225 or 135 x θ = ° ° CCW from the x axis). The point of application, 3 , r v can be found from the second condition, 3 3 3 3 ˆ ˆ ˆ ˆ ( )( ) r F x i y j Fi F j × = + = v v v v 3 3 1 1 2 2 ˆ ˆ ˆ ˆ ( ) (1 m) (1 m) . x y F k r F r F j Fi Fk + =  × × = × = v v v v Thus, 3 3 1 m, x y + = or the line of action of 3 F v passes through the point of application of 2 F v (the point (0,1m)). Any point on this line is a suitable point of application for 3 F v (e.g., the point (0,1m)). (b) In this case, 1 2 F F =  v v so 3 0, F = v but 1 1 2 2 r F r F × + × = v v v v 2 1 2 ( ) r r F × ≠ v v v so 3 3 0. r F × ≠ v v Thus there is no single force that can be added to produce static equilibrium....
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This note was uploaded on 09/12/2009 for the course PH PH101 taught by Professor Prof.kim during the Spring '09 term at Korea Advanced Institute of Science and Technology.
 Spring '09
 Prof.Kim
 Physics, Static Equilibrium

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