A thin flexible gold chain of uniform linear density has a mass of 17.1 g. It hangs between two 30.0 cm long vertical sticks (vertical axes) which are a distance of 30.0 cm apart horizontally (x-axis), as shown in the figure below which is drawn to scale. 30 28 26 24 22 20 18 16 14 12 10 8 6 4 2 0 0 2 4 68 10 12 14 16 18 20 22 24 Evaluate the magnitude of the force on the left hand pole. 26 28 30
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- Please find the Grashof classification and the Barker classification of the two four-bars: (NOTE: distances are in non-dimensional units. Can use inches or cm) P1) L1=d=4, theta1=0 degrees, L2=a=1.6, theta2=60 degrees, L3=b=5.5, L4=c=4.5 P1) L1=d=2.5, theta1=0 degrees, L2=a=2, theta2=45 degrees, L3=b=1.5, L4=c=1.5 Please also draw the four-bars to estimate the unknown angles theta3 and theta4.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 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
- length L = 28 cm, form an inverted U. The vertical rods each have mass of 14 g; the horizontal rod has mass of 52 g. Where is the center of If the figure below, three uniform thin rods, each mass of the assembly (x, y)? cm3) 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.)QUESTION 4 Find the resultant vectors. This referes to Question 4 on the PDF v Horizontal 3ON and Vertical 60N A. 67.1N @ 116.6 below the horizon v Horizontal 3ON and Velucal -60N B. 67.1N @ 63.4 above horizon v Horizontal -30N and Vertical 60N C. 67.1N @ 243.4 from the horizon v Horizontal -30N and Vertical -60N D. 67.1N @ -63.4 below the horizon
- 3) A right circular cone of 30 mm diameter and 60 mm height is cut from a cylinder of 50 mm diameter and 120 mm height as shown in the figure. Find the position of the CG of the body from its base. 50 Ans CG of the given section is at (0mm, 62.87mm,0mm)Two students are playing catch outside. One throws the ball too high and overshoots his friend, hitting the streetlight - whoops! What is the height hB of the ball when it passes over player B? VA В hB ho A ha ha * dBc dAB dAB = 27 ft h4 = 8 ft dBC = 4 ft h. = 10.7 ft 0 = 29° What is the magnitude of v4? Enter your answer correct to two 2 decimal places. ft/s VA4.14. Calculate I,,Iy,Iyy for the symmetrics Z-section of Figure P4.8
- 16 - Find the center of gravity M (x,y,z) of the three-dimensional homogeneous ABCD wire in the figure by using the coordinate set. The dimensions of the wire are a= 24 cm, b= 36 cm. In which of the following options are the coordinates of the homogeneous wire in the x, y and z axes given correctly?A) (23.57 ; 11.43 ; 17.14)B) (23.57 ; 17.14 ; 11.43)C) (28.29 ; 13.71 ; 20.57)D) (20.61 ; 13.03 ; 9.08)E) (28.29 ; 20.57 ; 13.71)For the area below, evaluate 1. The (x,y) coordinates of the centroid (please mark your origin or reference point on the diagram) 2. The 2nd moments and product of area in centroidal-based xy (horizontal-vertical) coordinates 3. The principal 2nd moments of area, and the direction of the first principal axis relative to the horizontal (x) axis (anticlockwise positive) 50 75 30 150 60 50 15 Figure 1: Cross section (all dimensions in mm) Quantity Xc (please mark your origin on the diagram) Yc (please mark your origin on the diagram) Ix ly Ixv I₁ I₂ a1 (please show the direction) Value 30 UnitsCompare 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]