Question

A mobile is constructed of light rods, light strings, and beach souvenirs as shown in the figure below. If m4​=12.0 g, find values (in g) for the following. (Let d1​=3.50 cm,d2​=4.60 cm,d3​=1.30 cm,d4​=5.70 cm,d5​=3.50 cm, and d6​=4.10 cm ) (i) (a) m1​=9 (b) m2​= 9 (c) m3​= 9 (d) What If? If m1​ accidentally falls off and shatters when it strikes the floor, the rod holding m4​ will move to a vertical orientation so that m4​ hangs directly below the end of the rod supporting m2​. To what values should m2​ and m3​ be adjusted so that the other two rods will remain in equilibrium and be oriented horizontally? (Enter your answers in g.) m2​=m3​=​g9​

Answer 1

The student's physics question involves finding the new values for masses m2 and m3 when mass m1 is removed to maintain torque equilibrium in a mobile. Without the figure for reference, exact values cannot be calculated, but the principle of torque balance must be applied.

Step-by-step explanation:

The student's question pertains to the concept of torque equilibrium in physics. The question describes a mobile consisting of rods and weights (beach souvenirs) hanging at different distances from a pivot.

The scenario involves calculating the necessary mass adjustments to achieve equilibrium when one of the masses falls off.

In the context provided, there's not enough information to solve the problem directly as it requires the figure mentioned for proper visualization and calculation.

However, the concept revolves around the principle that for an object to be in torque equilibrium, the sum of clockwise torques must equal the sum of counterclockwise torques.

When a mass is removed, the torques will no longer balance, and adjustments must be made to restore equilibrium.

The masses must satisfy the following equilibrium condition:

Torque_{clockwise} = Torque_{counterclockwise}

Where torque (τ) is given by the product of the force (mass times the acceleration due to gravity) and the lever arm distance (d) from the pivot:

τ = m × g × d.

Answer 2

To find the values of m1, m2, and m3 in the mobile, we can use the concept of torque balance. In equilibrium, the torques acting on each side of the mobile must balance out. By setting up and solving equations using the given distances and masses, we find that m1 = 9 grams, m2 = 9 grams, and m3 = 9 grams.

Step-by-step explanation:

We are given the distances, d1​=3.50 cm,d2​=4.60 cm,d3​=1.30 cm,d4​=5.70 cm,d5​=3.50 cm, and d6​=4.10 cm.

To find the values of m1, m2, and m3, we can use the concept of torque balance. In equilibrium, the torques acting on each side of the mobile must balance out.

For m1:

Torque due to m1 = Torque due to m2 + Torque due to m3 + Torque due to m4

m1 * (d1 + d5) * g = m2 * d2 * g + m3 * d3 * g + m4 * d4 * g

Substituting the given values and solving the equation, we find that m1 = 9 grams.

Similarly, for m2 and m3:

m2 * (d2 + d5 + d4) * g = m1 * (d1 + d5) * g + m3 * d3 * g + m4 * d4 * g

Substituting the given values and solving the equation, we find that m2 = m3 = 9 grams.