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- A.) It is the boundary condition for fixed support Choices: Both choices Y=0 Y’=0 Either of the choices B.) It is a type of simple stress acting parallel to an area Choices: Flexural stress Shear stress Axial stress Bearing stress C.) For concentric row of bolts in a coupling connection, the shear deformations in the bolts is proportional to the: Choices Modulus of rigidity Diameter of bolt circle Radial distance Number of bolts note: please help us in the word meaing sample problems thank youQ1/ Find the two components of the force (150 N) if a= 40degrees, as shown in the figure below FY F= 150 N a= 40 FX (FX= 114.9 N) AND (FY= 96.4 N) O (FX= 96.4 N) AND (FY= 114.9 N) O (FX= 130 N) AND (FY= 90 N) O (FX= 90 N) AND (FY= 130N) O3m Find the rectangular representation of the force F, given that its magnitude 240 N. * F = -135.76 i+ 169.71 j– 101.82 k F = 135.76 i + 169.71 j+ 101.82 k F = 135.76 i - 169.71 j+ 101.82 k F = -135.76i+ 169.71 j+ 101.82 k
- Q1: Draw just the Free Body Diagram of the following figures. Figure (1) Figure (2) Figure (3) 30° T45° 45° 60° 30 B 1 kN Figure (4) Figure (5) 26 kN 250 N 30 C 13 12 30% 0.15 m B 0.4 m 2 m 4 m 0.2 m ||10 / 36 110% Q22](problem 2/28: p37) (¿ãt2) Determine the resultant `R" of the two forces Shown in fiqure, and find the angle o" between R and A- 10 800 N the x-axis . 900 N 25 4" Ane R=10379 N . 0 =66-8 Q23]( problem 2/25 's p37) Exemple201-2o At What angle x must the Type here to searchQI5: Delermhe The resultant o. The forces shoan in figure () 4 KN 6.A 48KN CAKN 2 KN 54 KN Fy-(15)
- The following diagram corresponds to questions 1 to 3. A beam rests on two sharp edges as shown on the image. The beam has a length of 7,50 m and a mass of 4,25 kg. Object 1 has a mass of 1,50 kg; object 2 has a mass of 2,35 kg. Point P is 0,50 m form the center of the beam. d 2 Pl CG 1. How much is the torque done by object 1 around point P? A. 55,2 N*m B. 7,36 N*m С. 62,5 N*m D. Object 1 doesn't exert a torque.The following diagram corresponds to questions 1 to 3. A beam rests on two sharp edges as shown on the image. The beam has a length of 7,50 m and a mass of 4,25 kg. Object 1 has a mass of 1,50 kg; object 2 has a mass of 2,35 kg. Point P is 0,50 m form the center of the beam. CG 1. How much is the torque done by object 1 around point P? A. 55,2 N*m B. 7,36 N*m C. 62,5 N*m D. Object 1 doesn't exert a torque. 2. How much is the torque done by the force of gravity of the beam around point P? A. 20,8 N*m N-m ך177 .B C. 156 N*m D. 313 N*m 3. If you needed to cancel the nomal forces of the two objects, where you should place object 2? The axis of rotation is point P. A. 3,30 m from point B. 9,18 m from point P C. 5,69 m from point P D. 3,62 m from point P(٢٦+ * Find the two forces Fbc & Fac as :shown in the figure below Fab=30 N 110 30/Fbc (Fbc=45.18 N and Fac= 58.8 N) O (Fbc=38.56 N and Fac= 56.38N) O (Fbc=75.18 N and Fac= 78.8 N) O ! Fac
- Compare the magnitudes of the equilibrant vectors measured from the experiment with those obtained from the graphical and component methods. Example: A: 200 g 60° above +x axis B: 300 g 45° above -x axis C: 400 g 30°below -x-axis A, A cos a = 1.96 N x cos 60° = 0.98 N B₂B cos b = 2.94 N x cos 135º = -2.08 N C, C cos g = 3.92 N x cos 210°= -3.39 N R₂-A, + B + C₂ = -4.49 N A, A sin a = 1.96 N x sin 60° = 1.70 N By B sin b = 2.94 N x sin 135º = 2.08 N C, C sin g = 3.92 N x sin 210° = -1.96 N Ry= Ay+ By + Cy = 1.82 N Questions: I: (a) A: 200 g along +x axis B: 100 g 45° above -x axis A₂ = A cos a = B₂ =B cos b = R₂-A₂+ B₁₂= A₂ = A sin a = B, = B sin b = R₂ = A + B₂ = R=(R₂. R₂):_ Quadrant R = √ (R₂²+R₂²) = Direction:q=tan [R, /R.] (c) A: 100 g along -y axis B: 200 g along -x axis A = A cos a = B = B cos b = R₂-A₂+ B₂ = A = A sin a = B, B sin b = R=A, +B₂ = R=(R₁, R₂): Quadrant R = √ (R₂²+R₂²) = Direction:q tan¹ [R, /R]S00 Ib In the fig. shown, compute the ff: (16-18) the resultant using cosine law (force polygon) 60 R = 35 (19-20) the angle of the R measured 500 lb cW from the x- axis.F = FA i = -(d,7 + d,j + d,k) d Fd Fd y Fd , F, = d F, = d d Example : The tension in the guy wire is 2500 N. Determine: a) components F, F, F of the force acting on the 80 m - 40 m bolt at A, 30 m b) the angles q, 9, 9 defining the direction of the force ON: and B, determine the