Practice Problems #7: COG and Angular Kinematics

Complete the table shown below (you have to do another table) by mentioning the joint actions, corresponding planes and axes of motion of a barbell back squat (use the figure as a reference).

Flexion/extension (Plantarfelxion/dorsiflexion)occurs on the transverse axis Plane: Sagittal

A barbell is loaded with two 20 kg plates on its right side and two 20 kg plates on its left side. The barbell is 2.2 m. long, and its unloaded mas is 20 kg. The two 20 kg plates on the right side are locked in place at 35 cm and 40 cm from the right and of the bar. The two 20 kg plates on the left side have slipped. One is 30 cm from the left end of the bar, and the other is 20 cm from the left end of the bar. Assuming that the axis of rotation is on the far right side of the barbell, where is the center of gravity of this barbell with the four plates (Show variables known, unknown, equation to be used, and solve it step by step)?

Variables Known: Barbell: 2.2 m long, Unloaded barbell mass: 20 kg, both sides of barbell have 2 (20 kg) plates, and the axis of rotation is on the far right side. Unknown Variables: Where is the center of gravity of the barbell with all four plates? Equation being used: m = mass, x = position, n = total # of particles x = summation(m1x1...)/m1 Solve: x = (4kgm+6kgm+22kgm+36kgm+37kgm) / (100 kg) = 1.05 m from left end

In lacrosse, an attacker's stick is 101.6 cm long. The stick is 350 g and due to the weight of the basket, has a center of gravity 10 cm toward the basket from the center. Presume the player's right hand is centered, and consider it the axis of rotation. The ball has a mass of 140 g and sits in the basket 51 cm from the axis. The player's left hand also provides a downward force of 3 N at a point 40 cm from the center in the opposite direction. Including the weight of the stick, the ball, and the force of the left hand, what is the stick's new center of gravity (Show variables known, unknown, equation to be used, and solve it step by step)?

Weight of objects must be expressed in N (e.g. 0.14 kg * -10 m/s/s = 1.4 N) Here is how to solve it: W1 = .14 * -10 = -1.4 r1 = -.51 m W2=.350 * -10 = -3.5 r2 = -.1 m W3 = -3N r3 = .40 m Σ(W × r) = (ΣW) × rcg Σ(W × r) = (0.14kg)(-10 m/s/s)(-0.51m) + (0.35kg)(-10 m/s/s)(-0.1m) + (-3N)(0.40m) Σ(W × r) = 0.714 + 0.35 - 1.2 = -0.136 -0.136 = (-1.4 - 3.5 - 3) × rcg -0.136 = (-7.9) × rcg 0.017 m = rcg New CoG is 1.7 cm from the axis, in the direction of the back hand