The 66 kg man's hands in Figure 9-56 are 36 cm apart. His CG is located 90% of the distance from his right hand toward his left. Find the force on each hand due to the ground.
a)the man's left hand
b)the man's right hand 2
To find the force on each hand due to the ground, we can use the principle of equilibrium. In equilibrium, the sum of all the forces acting on an object is zero.
Let's break down the problem:
Step 1: Determine the distance between the man's hands.
In the problem, it is stated that the man's hands are 36 cm apart.
Step 2: Calculate the distance of the man's CG from his right hand.
The problem states that the man's CG is located 90% of the distance from his right hand toward his left. Since the total distance between his hands is 36 cm, 90% of this distance gives us (0.9 * 36) cm = 32.4 cm. Therefore, the distance of the man's CG from his right hand is 32.4 cm.
Step 3: Calculate the force on each hand due to the ground.
Let's assume the force on the man's left hand is F_left and the force on the man's right hand is F_right.
Since the man is in equilibrium, the sum of the forces horizontally and vertically must be zero:
Horizontal forces: There are no horizontal forces acting on the man, so the sum of the horizontal forces is zero.
Vertical forces: The sum of the vertical forces is equal to the weight of the man, which can be calculated as the product of his mass (66 kg) and acceleration due to gravity (9.8 m/s^2).
Now, let's consider the moments (torques) about a point in the air, which we can choose to make the calculations easier:
Taking the moments about the man's left hand:
The weight of the man acts vertically downward at the location of the CG. The distance between the CG and the left hand is (36 cm - 32.4 cm) = 3.6 cm. Therefore, the moment produced by the weight of the man is (66 kg * 9.8 m/s^2) * (0.036 m) = 22.1784 N.m (or equivalently, kg.m^2/s^2).
Now, since the man is not rotating about any point (in equilibrium), the sum of the moments must be zero. Therefore, the force on the man's left hand (F_left) multiplied by the distance between the left hand and the point where we took the moments (which is 0), is equal to the moment produced by the weight of the man. This gives us:
F_left * 0 = 22.1784 N.m
Therefore, the force on the man's left hand is zero.
Taking the moments about the man's right hand:
The distance between the CG and the right hand is 36 cm - 3.6 cm = 32.4 cm. Therefore, the moment produced by the weight of the man is (66 kg * 9.8 m/s^2) * (0.324 m) = 209.5632 N.m.
Again, since the man is in equilibrium, the sum of the moments must be zero. Therefore, the force on the man's right hand (F_right) multiplied by the distance between the right hand and the point where we took the moments (which is 0), is equal to the moment produced by the weight of the man. This gives us:
F_right * 0 = 209.5632 N.m
Therefore, the force on the man's right hand is zero.
Hence, the force on both the man's left hand and right hand due to the ground is zero.