For the beam in shown in the figure, determine the vertical displacement in C & and the rotation in A using Castigliano's Theorem. E = 200 GPa; 1= 150X10° mm². A 8 kN/m -4 m 20 KN C -4 m- B
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- 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.A square plate with sides 1.5 m in length can rotate around an axle passing through its center of mass (CM) and perpendicular to its surface (see figure below). There are four forces acting on the plate at different points. The rotational inertia of the plate is 16 kg - m². Use the values given in the figure to answer the following questions. (Assume 0 = 40°. Express your answers in vector form.) 30.0 N 60.0 N CM 20.0 N 40.0 N (a) What is the net torque acting on the plate?The triangular plate is fixed at its base, and its apex A is given a horizontal displacement of 5 mm. Suppose that a= 640 mm
- You are required to calculate the second moment of area about the x-axis for the figure below. a mm b mm a mm a mm One method for calculating the second moment of area is to break up the shape into three parts as shown below. Calculate the second moment of area of segment 2 if: a = 30.67 mm b = 21.14 mm |_x = mm^4 a mm ww eThe displacement vectors A and B shown in the figure below both have magnitudes of 3.10 m. The direction of vector A is 0 = 23.0°. (a) Find A + B graphically. magnitude m direction o counterclockwise from the +x axis (b) Find A – B graphically. magnitude direction o counterclockwise from the +x axis (c) Find B - A graphically. magnitude direction o counterclockwise from the +x axisF = F cos # F = F + F F F sin 8. # tan Exampie (1):Find the two components of the force (100 X) if: 0 = 30", 120 270° as shown in figure. F = 100 N F = 100 N 6=30 0 =120° 0 =270 0 = 30 0 =60 F = 100 N lution:
- Show conv D +z axis 01) Given the figure, AB=5m, BC= 8m and AG= 7m and pt 0 is origin. Determine the following: 250 N B *y axis 1) What is magnitude of R and Cr in pt G? 20 kN m 1689 Ibt (Unit must be in N and kN-m) 1269 pounds 2) What is the magnitude of R and Cr in pt 0? +x axis (Unit must be in N and kN-m) 3) Unit Vector of R in the system. 4) Direct Cosines of Cr in pt 0.3) A point mass is attached to one end of a beam levered against a spike which acts as the pivot. The point mass weighs 250N and the beam has a uniform mass of 10kg and is 8m long. Parts a and b Part c Fapplied 3 of 6mov 1m 8m Krist 0 = 60° Scale 250 N Scale a. If a scale placed beneath the point mass reads 250N, what is the magnitude and direction for the force at the pivot point? Direction: North, 90° from x-axis Magnitude to k* 250N = 2500 N b. For the same scenario, how far must the pivot point be located relative to the left edge of the beam? (Hint, both the boxes weight and the force from the scale act at the farthest right edge of the beam.) www.acta 0672nos onl c. The pivot is then moved to be 6m from the left edge of the beam which now makes a 60 angle with the floor. How much downward force would be needed at the left-most edge of the beam to reduce the reading on the scale to zero? (Hint: the beam is still in static eq. when the scale reads zero.)у- ахis F4 F5 X- axis BC F1 х- ахis y - axis F1 F2 Figure (1) Figure (2) 250 N 36.870 D F2 60° 0.5 m 1 m B 0.2 m 1 m 300 NE F1 1 m Figure (3) Figure (4) Q1] Answer the following questions: For figure (1): The object is subjected to two forces F1 = 25 N, and F2 = 50 N. Set x = 30°, 0 = 45°, and ß = 30°. 1- What is the magnitude of the resultant of these two forces? a) 39.4 N b) 80.1N c) 41.6 N d) 69.8 N e) 71.4 N f) 46.1 N g) None of them 2 - What is the direction of the resultant force measured counterclockwise from the positive x-axis? a) 14. 3° b) 392.6° c) 9.2° d) 322.9° e) 6.8° f) 344.2° g) None of them 500 N
- 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.Solve the preceding problem for the element shown in the figure.Review Learning Goal: To use the vector cross product to calculate the moment produced by a force, or forces, about a specified point on a member. Part A - Moment due to a force specified by magnitude and endpoints The moment of a force F about the moment axis passing through O and perpendicular to the plane containing O and F can be expressed using the vector cross product, Mo =r x F. In a properly constructed Cartesian coordinate system, the vector cross product can be calculated using a matrix determinant: As shown, a member is fixed at the origin, point O, and has an applied force F, the tension in the rope, applied at the free end, point B. (Figure 1) The force has magnitude F = 180 N and is directed as shown. The dimensions are ¤1 = 0.350 m, x2 = 1.90 m, y1 = 2.30 m, and z1 = 1.20 m. What is the moment about the origin due to the applied force F? i j M =rx F =|rz k Express the individual components of the Cartesian vector to three significant figures, separated by commas. ry F F,…