the welght of the red 70N 0.8 m SON 1.3 m
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- Six forces are acting on the mesh as shown in the figure below. Find the total torque produced by the given forces at points (a) A and (b) B. 255N + 1.0 ml 112 N 1.0nt A 384 N 429 N B 328 N 183NBoth the magnitude and the direction of the force on a crankshaft change as the crankshaft rotates. Find the magnitude of the torque on the crankshaft using the position and data shown in the figure, where F = 1100 lb. ft-lb 60° F 0.16 ft --L-------Q; Let force F, = -i-}, 7 = } , Â--21-4j and Fu= 2i are acting at OC0,0), A(20), BL0,2) and C(3,2) xespectively. iFind sum of torques of all the forces about O.
- 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 rectangular framework is subjected to the following non-concurrent system of forces: a 30lb- force below the origin, a 20 lb-force,3ft above the origin and making an angle of30°N of E, a 100lb-force with a horizontal distance of 3ft and a vertical distance of 4ftfrom the origin and making an angle of 45°NW , a 60lb -force with the same distance like the previous force but along the positive x-axis, and a 100lb- force 4ft tothe right of the origin and making an angle of 60° S of W . Determine the magnitude and direction of the resultant, as well as its moment arm relative to the origin.A rod is 0.70 m long and lies along the x-axis, with one end at the origin. A force of 8.0 N is applied at the point x = 0.50 m, and is directed 30° above the x-axis. What is the torque on the rod? O 0.80 N-m ○ 2.0 N.m O 4.8 N.m O 8.0 N-m O 10 N.m Save for Later Submit Answer
- Six forces are acting on the mesh as shown in the figure below. Find the total torque produced by the given forces at points (a) A and (b) B.:Q81 The moment M for the force the figure shown below is 250 N 153 -200 mm 30 mm 46.5 N.M O 46.6 N.M O 46.4N. M of L isConsider the square plate ABCD shown in the figure to the right whose rotation is about an axis through its center at point O and perpendicular to the plate. The forces are applied at different points on the plate: Fi - 10.0 N is applied at point B, F: - 7.50 Nis applied at point D and Fs-9.00 N is applied at the midpoint between the side BC. (a) Calculate the magnitude the torque due to each force. (b) Determine the magnitude and direction (in terms of rotation) of the net torque on the rhombic plate. (c) lf a fourth force is to be applied at point A, what must be the magnitude and direction of that force such that it is minimum and that the plate attains rotational equilibrium? 143.1 30 cm
- (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 ! FacHey, please reply as soon as posible, thanks! The arms AB and BC of the desk lamp shown in the figure are contained in a plane that makes an angle of 30º with the xy plane. The following forces and torques are applied: •A force F1, parallel to the y axis, applied at the point middle of bar AB. •A force of 8 N at point C, raised 45º above line CD which is parallel to the z axis. •A torque of 10 Nm at point B, which is parallel to line OE. F1 = 9N AB = 360 mm BC = 266 mm 1. Calculate and indicate the Force-Pare equivalent system in A 2. Assuming that the system is in equilibrium and that the support in A no allows movement in either direction (both displacement and rotation), indicate the reaction forces and moments acting at point A.am of Mechanics/Firs X E Monthly Exam of Mechanics/Firs x PQLS ZSh4aZJdw9EBqROOfLVx M7KUGF6 9bgBbd-YGOXA/formResponse * For the force shown in figure, find the resultant (R) by component method F=10 kN 0=30° R=5 kN R=10 KN O R=15 kN O R=60 KN For the forces shown in figure, find just (EFx) components F=4 kN F=2 kN 0-30 A=300