1. LeBron uses his bicep and forearm to lift and hold a 6.0 kg lead ball still n
ID: 1779673 • Letter: 1
Question
1. LeBron uses his bicep and forearm to lift and hold a 6.0 kg lead ball still n his hand. All forces are located ith distances from his joint as shown in the graph below. The weight at the center of the mass of his forearm is 12 N. (and let's call it FE on Forearm in this problem) His bicep muscle has a force Fs exerted on his arm. Find the force FB. (Hint: all three forces in the graph should be balanced out with the 4th vertical force at the elbow joint. But we will ignore that 4th force in this problem.) FEon Form 12N mg=6.0 kg 0.05m elbow joint 0.16 m 0.35 m 2. Several 100 g identical marbles are fixed on on each massless rod. Each rod is fixed by a pivot on the horizontal plane and is rotated with a constant angular velocity. The length L of each massless rod is 1.20 m. Find the rotational inertia on the entire rigid body, and rank the tangential speed (label them as v1, V2, v3.. .etc.) from the greatest to smallest for the (i) the left rod, and (ii) the right rod. pivot pivot L/4 L/4 L/4 L/4 L/3 L/3 L/3Explanation / Answer
question 3)
Figure 1: First we need to find the angle which the forces F1 and F2 make with the x axis and the - x axis respectively
For F1: tan 1 = F1y/F1x = 1/2
1 = 26.56 degrees
Similarly 2 = 26.56 degrees
Total force in x direction = F1 cos 1 + F2 cos 2
F1 = (22 + 12) = 51/2 = 2.23 N = F2
F1x = 2.23 cos(26.56) - 2.23 cos(26.56) = 0
F1y = 2.23 sin(26.56) + 2.23sin(26.56) = 1.994 ~ 2 N
Therefore the particle is not in equilibrium in the y direction because a net force of 2 N is acting on it in that direction. A vector pointing in the y direction, starting from origin and its head ending at 2 can be drawn to show the same.
Figure 2
Following the above logic
F1 makes and angle of tan = 1/2 with the y direction, 1 = 26.56 w.rt to y direction
F2 makes an angle of 2 = 26.56 with the negative y direction.
F1=F2 = 2.23 N (as calculated above)
total force in x direction = F1sin(26.56) + F2 sin(26.56) + 2 -2 = 1.994 N ~ 2 N
total force in y direction = F1 cos(26.56) - F2 cos(26.56) = 0 N
The particle is not in equilibrium in the x direction. A vector starting at the origin and ending at 2 in the x direction can be drawn to represent the same
Figure 3
Using the above method, we can see that F1 makes and angle of 45 (tan = 1/1) with x axis, F2 makes an angle of 45 with the -x axis.
F1 = F2 = (12 +12) = 1.414
Total force in x axis = 2 + F1 cos(45) - F2cos(45) = 2 N
Total force in y axis = F1 sin(45) + F2sin(45) -F3 = 1+1 -1 = 1 N
The particle is experiencing a force in x as well as y direction. Hence not in equilibrium
Fx = 2 N, Fy = 1 N
A vector starting at origin, and ending at (2,1) represents the net force vector.
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