chap 12 - STATIC EQUILIBRIUM 12 2 EXERCISES Section 12.1...

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12.1 STATIC EQUILIBRIUM EXERCISES Section 12.1 Conditions for Equilibrium 14. (a) ∑= + − − + = v Fi j i j j i ( ˆˆˆˆ ˆ ). 2223 0 N (b) () ˆ ( ˆˆ )( ˆ ) ∑= × ++ × −+ r τ i iij i ij 0 22 2 2 3 7 [( ˆ ˆ ) ˆ ] ij j ⋅= Nm ˆ . 437 0 +− ⋅ = k 15. INTERPRET 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. DEVELOP The three forces are r Fij 1 =+ N at point (,) ( ) , xy = 2m ,0m r j 2 23 =− − N at , 1m and r Fj 3 1 = ˆ N at . 7m ,1m We find the torques due to these three forces around points 3m ,2m and , using r r r τ || . rF 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: rr r r =− . applied pivot EVALUATE For point , ,2m the torque due to F 1 is r r 11 02 × = + ( ( ) ˆ rrF i applied pivot m ˆ ) ˆ ( ji j k mN Nm 1 ×+ = − − 120 220 2 r ) −= 42 kk Nm Nm Similarly, r 2 8 = ˆ k Nm and r 3 10 ˆ k Nm. The sum of these three is r total k =+− = ˆ . 2 8 10 0 Nm Around point (–7 m, 1 m), the torques are r 1 20 = ˆ k Nm, r 2 20 ˆ k Nm, and r 3 0 = . The sum of these three is also r total = 0. ASSESS 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: vv v FF F 31 2 and v v v v v v rFrF 33 ×= −×+× . (a) In this case, v j r j i 2 == = ˆ ,( , m) , 2 and v rj 2 1 = ˆ . m Thus, v i j 3 + ( ), which is a force of magnitude 24 5 F , ° down into the third quadrant ( θ x ° 225 135 or CCW from the x axis). The point of application, v r 3 , can be found from the second condition, v v rF x iy jF iF j 3 3 ×= + − − = ( ) ˆ = −×−× =− × = F k rFrF i 1122 01 v v v v m( ) ˆ . Fk Thus, −+ = 1m, or the line of action of v F 3 passes through the point of application of v F 2 (the point ( , )). 01m Any point on this line is a suitable point of application for v F 3 (e.g., the point ( , (b) In this case, 12 so v F 3 0 = , but v v v v 112 2 ×+×= v rr F 21 2 0 −×≠ so v v 0 ×≠ . Thus there is no single force that can be added to produce static equilibrium. 17. INTERPRET The problem asks for a set of conditions that must be met for the body to be in static equilibrium. This means that all the external forces and torques must be zero. DEVELOP Static equilibrium demands that (see Equations 12.1 and 12.2) Σ ΣΣ r r r r F i ii i = = 0 0 12
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12.2 Chapter 12 Since all of the forces lie in the same plane, which includes the points O and P , there are two independent components of the force condition (Equation 12.1) and one component of the torque condition (Equation 12.2). EVALUATE (a) Taking the x axis to the right, the y axis up and the z axis out of the page in Figure 12.12, we have: 0 0 0 12 3 24 =∑ =− + + =− + FF F F F x y zO sin cos () φ τ + LF 23 sin cos φφ (b) The equation for torque about point P is 0 21 2 1 2 + s i n( ) τφ zP L L F The lever arms of all the forces about either O or P should be evident from Fig. 12.12.
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chap 12 - STATIC EQUILIBRIUM 12 2 EXERCISES Section 12.1...

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