Problem 86 - Fundamentals of Physics

Question

The tension in a string holding a solid block below the surface of a liquid (of density greater than the block) is \(T_{0}\) when the container (Fig. \(14-57\) ) is at rest. When the container is given an upward acceleration of \(0.250 \mathrm{~g}\), what multiple of \(T_{0}\) gives the tension in the string?

Step-by-Step Solution

Answer

The tension in the string is 1.25 times \(T_0\).

Step 1: Write down the known values

We have a block of mass \(m\) and the acceleration due to gravity, \(g\). The container is given an upward acceleration of \(0.250g\), which means its total effective acceleration becomes \(a = g + 0.250g = 1.250g\). The tension at rest is \(T_0\).

Step 2: Establish the tension equation when at rest

When the block is at rest, the forces acting on it are the gravitational force downward, \(mg\), and the buoyant force, which we will call \(B\) (an upward force). The tension in the string at rest must balance these forces, so we have:\[ T_0 + B = mg \]Or solving for \(T_0\):\[ T_0 = mg - B \]

Step 3: Establish the tension equation when accelerating

With the container accelerating upwards, the effective weight of the block increases to \(m(1.250g)\), due to the additional acceleration. The tension \(T\) must now balance this increased effective weight and the buoyant force:\[ T + B = m(1.250g) \]

Step 4: Solve for the new tension

Subtracting the buoyant force \(B\), we have:\[ T = m(1.250g) - B \]

Step 5: Relate the new tension to original tension

We already have \(T_0 = mg - B\). Divide \(T\) by \(T_0\):\[ \frac{T}{T_0} = \frac{m(1.250g) - B}{mg - B} \]Replace \(m(1.250g)\) and \(mg\) in the ratio:\[ \frac{T}{T_0} = \frac{1.250mg - B}{mg - B} \]Factor out \(mg\):\[ \frac{T}{T_0} = \frac{1.250(mg - B) + 0.250B}{mg - B} \]

Step 6: Simplify the expression

Using \(T_0 = mg - B\), replace \(mg - B\) in the expression:\[ \frac{T}{T_0} = \frac{1.250T_0 + 0.250B}{T_0} \]Assume buoyant force \(B\) changes insignificantly with additional gravity (as commonly water density doesn't change significantly with pressure), so:\[ \frac{T}{T_0} = 1.250 \]

Step 7: Conclusion: State the final result

The tension in the string when the container is accelerated upward at \(0.250g\) is 1.25 times the tension \(T_0\) when at rest. Thus, the multiple of \(T_0\) that gives the new tension is 1.25.

Key Concepts

Tension in a String

When we talk about the tension in a string, we're referring to the force exerted by the string when it is used to hold or pull an object. In this scenario, a block is submerged in a liquid and held by a string. The tension in the string is a crucial factor because it counters other forces acting on the block, such as gravity and buoyancy.

- **At rest:** When everything is stationary, the tension equals the block’s weight minus any buoyant force from the liquid. Symbolically, we express this as \(T_0 = mg - B\), where \(mg\) is the gravitational force on the block and \(B\) is the buoyant force.

• The tension here keeps the block from sinking further or floating up, depending on the liquid’s density.

- **In motion:** With upward acceleration, more force is needed in the string, increasing the tension. This accounts for both gravity and the upward pull from acceleration, expressed as \(T = m(1.250g) - B\). By comparing tensions at rest and during motion, we see that tension increases by a factor (in this case, 1.25).

Effective Weight

Effective weight is a concept that describes how much an object seems to weigh, considering all acting forces and acceleration. It is different from the actual weight, which is simply \(mg\), the gravitational force.

• In non-accelerated states, effective weight is the same as actual weight.
• When moving, effective weight changes due to additional accelerative forces.

When the block experiences an upward acceleration of \(0.250g\), its effective weight is no longer just \(mg\). Instead, it becomes \(m(1.250g)\), because it feels heavier due to the upward force. This altered weight influences how much tension the string must exert to keep balance.

Upward Acceleration

Upward acceleration refers to any increase in motion that moves against the direction of gravity. In this problem, the container is moving upwards with an acceleration of \(0.250g\).

• This means that every part of the system feels an extra force upward.
• The block not only deals with gravity but also this additional upward pull.

Here's the key effect: the block's need for more string tension to hold it steady. The combined effect of this upward acceleration and gravity modifies the total pulling force on the string. The formula \(a = g + 0.250g = 1.250g\) captures this larger, effective gravity-like force experienced as a result of the acceleration.

Gravitational Force

Gravitational force is the pull that Earth exerts on objects, calculated as \(mg\) (mass times acceleration due to gravity). It is the fundamental downward force affecting all objects near Earth’s surface.

• It acts equally and downward to ensure that items have weight.

In the context of a submerged block, gravitational force works together with buoyant force. The block is pulled down by gravity, and held up by the buoyancy of the liquid. The tension in the string then compensates for the difference to keep the block suspended. In dynamics, any acceleration, such as the upward acceleration in this problem, modifies the effective influence of gravity, but the gravitational force remains a constant downward force. It forms a baseline in calculating both effective weight and the needed tension adjustments.