To solve this problem, let's approach it with the principles of equilibrium for forces and moments (torques).
### Step 1: Understanding the Problem
- The plank is 2.00 meters long.
- The mass of the plank is 16.0 kg.
- The force of gravity acts at the center of gravity, which is at the midpoint of the plank (1.00 meter from each end).
- One hand is pushing down on one end with a force [tex]\( F_1 \)[/tex].
- The other hand is holding the plank up at a point 0.750 meters from the end with a force [tex]\( F_2 \)[/tex].
- We need to find the magnitudes of the forces [tex]\( F_1 \)[/tex] and [tex]\( F_2 \)[/tex].
### Step 2: Calculate the Weight of the Plank
First, calculate the weight of the plank using the formula:
[tex]\[ \text{Weight (W)} = \text{mass} \times g \][/tex]
where [tex]\( g \)[/tex] (acceleration due to gravity) is approximately [tex]\( 9.8 \, \text{m/s}^2 \)[/tex].
[tex]\[ W = 16.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 156.8 \, \text{N} \][/tex]
### Step 3: Set Up the Moment Equations
We choose the point where the hand is holding the plank (0.750 meters from the end) as the pivot point for calculating moments. The sum of the moments about this point must be zero for rotational equilibrium.
[tex]\[ \sum M = 0 \][/tex]
The moment due to [tex]\( F_1 \)[/tex] (applied at the end of the plank, 2.00 meters from the pivot) is:
[tex]\[ M_{F_1} = F_1 \times 2.00 \, \text{m} \][/tex]
The moment due to the weight of the plank (acting at the center of gravity, 1.00 meters from the pivot) is:
[tex]\[ M_{\text{weight}} = W \times (1.00 \, \text{m} - 0.750 \, \text{m}) = 156.8 \, \text{N} \times 0.250 \, \text{m} = 39.2 \, \text{Nm} \][/tex]
Setting up the equation for rotational equilibrium:
[tex]\[ F_1 \times 2.00 \, \text{m} = 39.2 \, \text{Nm} \][/tex]
[tex]\[ F_1 = \frac{39.2 \, \text{Nm}}{2.00 \, \text{m}} = 19.6 \, \text{N} \][/tex]
### Step 4: Set Up the Vertical Force Equilibrium Equation
For vertical equilibrium, the sum of the vertical forces must be zero:
[tex]\[ F_1 + F_2 = W \][/tex]
Substituting the values we know:
[tex]\[ 19.6 \, \text{N} + F_2 = 156.8 \, \text{N} \][/tex]
[tex]\[ F_2 = 156.8 \, \text{N} - 19.6 \, \text{N} \][/tex]
[tex]\[ F_2 = 137.2 \, \text{N} \][/tex]
### Conclusion
The magnitudes of the forces are:
- [tex]\( F_1 = 19.6 \, \text{N} \)[/tex]
- [tex]\( F_2 = 137.2 \, \text{N} \)[/tex]
Therefore, the magnitudes of the forces [tex]\( F_1 \)[/tex] and [tex]\( F_2 \)[/tex] are:
[tex]\[ F_1 = 19.6 \, \text{N} \][/tex]
[tex]\[ F_2 = 137.2 \, \text{N} \][/tex]
