There are three masses, M1 = 1 kg, M2 = 3 kg, and M3 = 5 kg.  M1 is located at (0,0); M2 is located at (2, 0); and M3 is located at (1, 1).  (All positions are in meters.)  Find the center-of-mass of this system of masses, with respect to the origin.  Express your answer in unit vector notation, i.e., rcm = xcmi + ycmj where r, i, and j are vectors.

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To keep the calculations fairly simple, but still reasonable, we shall model a human leg that is 92.0 cm long (measured from the hip joint) by assuming that the upper leg and the lower leg (which includes the foot) have equal lengths and that each of them is uniform. For a 70.0 kg person, the mass of the upper leg is 8.60 kg, while that of the lower leg (including the foot) is 5.25 kg. Find the location of the center of mass of this leg, relative to the hip joint, if it is (a) fully extended, and (b) bent at the knee to form a right angle with the upper leg.

The position of a particle according to the appropriate coordinate set can be vectorially specified in meters as r = i(sin2πt) -16j(1-e-t/4) + kt2/4 where t is the movement time of particle in seconds after passing its origin. If the mass of the particle is 2,4 kg, calculate the x component of its R force and the magnitude of its resultant force R at time t = 4s.

How do you find the the center of mass of a lamina whose density is given by p(x, y) on a region R?? Here m is the mass, M, is the moment in the x- axis and My is the moment in the y- axis М, М, Center of mass is located at т m The coordinates of the center are (M2, My) Find the moment M, and divide by the mass m The center of mass is located in the center of the universe. Find the moment M, and divide by moment My о ООО

Find the center of mass coordinates XCM y YCM of the arrangement of masses shown. Assume that the connecting bars are massless. (m= 3.0 kg; M = 1.5 kg; L = 2.0 m)

Determine the coordinates of the center of mass of the three-particle system below.  The masses are as follows, m1=3.0 kg, m2 = 4.0 kg and m3 = 8 kg.  (Use xcm= (x1m1+ x2m2+ ...xnmn)/(m1+m2+...mn) and a similar formula for y.)  Assume ys= 2 m and xs=2 m.

Three point masses ( 2 kg. 2 kg. 6 kg) shown in the figure are initially at rest on a frictionless horizontal floor. When the mass at the origin is subjected to a constant force F, it is observed that the center of mass (CM) of the three-mass system reaches point P with coordinates (4 m, 4 m) in 2 sec. Find the acceleration of the center of mass in unit vector notation. (Hint: after finding center of mass location, need to use kinematics) 2kg Y (m) 1X 4 2kg 4 メ (m)

A system of four particles moves along one dimension. The center of mass of the system is at rest, and the particles do not interact with any objects outside of the system. Find the velocity v4 of particle 4 at t1 = 2.75 s given the details for the motion of particles 1, 2, and 3. particle 1: mj = 1.33 kg, vi (t) = (7.41 m/s) + (0.277 m/s²) × t particle 2: m2 = 3.05 kg, v2 (t) = (8.05 m/s) + (0.567 m/s²) × t particle 3: m3 = 4.25 kg, v3 (t) = (6.33 m/s) + (0.195 m/s²) × t particle 4: m4 = 5.27 kg

4.2 Three point masses lie at the corners of an equilateral triangle whose sides have length 1.00 m. A 5.00 kg mass is located at the origin of the coordinate system, a 10.0 kg mass lies on the positive x-axis, and a 15.0 kg mass is in the first quadrant of the xy-plane. (a) Calculate the coordinates of the center-of-mass (CM) of the system. (b) How far is the CM from the origin, and what angle does it make with respect to the x-axis?