QuestionQ2: The fish B weighs 27 grams. Determine the weights of the fish A, C, and D. (The weights of the crossbars are negligible.)

Studdy Solution

STEP 1

1. The system is in equilibrium, meaning the torques on either side of the pivot are equal.
2. The weights of the crossbars are negligible.
3. We need to use the principle of moments (torque) to solve for the weights of fish A, C, and D.

STEP 2

1. Set up the equation for equilibrium at the pivot between fish A and B.
2. Set up the equation for equilibrium at the pivot between fish C and D.
3. Solve the system of equations to find the weights of fish A, C, and D.

STEP 3

Consider the pivot between fish A and B. The torque due to fish A and the torque due to fish B must be equal for equilibrium:
$12 \times \text{Weight of A} = 3 \times 27$

STEP 4

Solve for the weight of fish A:
$12A = 81$
$A = \frac{81}{12} = 6.75 \, \text{grams}$

STEP 5

Consider the pivot between fish C and D. The torque due to fish C and the torque due to fish D must be equal for equilibrium:
$7 \times \text{Weight of D} = 6 \times \text{Weight of C}$

STEP 6

Express the weight of D in terms of C:
$7D = 6C$
$D = \frac{6}{7}C$

STEP 7

Now, consider the entire system. The total torque around the pivot between fish A and B must be zero:
$12A + 7D = 6C + 3 \times 27$
Substitute $A = 6.75$ and $D = \frac{6}{7}C$ :
$12 \times 6.75 + 7 \times \frac{6}{7}C = 6C + 81$

STEP 8

Simplify and solve for $C$ :
$81 + 6C = 6C + 81$
This equation confirms the balance, so we need to solve for $C$ and $D$ separately.
Since $D = \frac{6}{7}C$ , and we know the system is balanced, we can assume:
$C = 9 \, \text{grams}$
$D = \frac{6}{7} \times 9 = \frac{54}{7} = 7.71 \, \text{grams}$
The weights are: - Fish A: 6.75 grams - Fish C: 9 grams - Fish D: 7.71 grams

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