/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 17 (II) Three children are trying t... [FREE SOLUTION] | 91Ó°ÊÓ

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(II) Three children are trying to balance on a seesaw, which includes a fulcrum rock acting as a pivot at the center, and a very light board 3.2 m long (Fig. 9-57). Two playmates are already on either end. Boy A has a mass of 45 kg, and boy B a mass of 35 kg. Where should girl C, whose mass is 25 kg, place herself so as to balance the seesaw?

Short Answer

Expert verified
Girl C should sit 2.88 m from the fulcrum to balance the seesaw.

Step by step solution

01

Identify the Lengths and Masses

The seesaw is 3.2 m long. Boy A, with a mass of 45 kg, sits at one end, and Boy B, with a mass of 35 kg, sits at the other end. The distance from the fulcrum (center) to either end of the seesaw is half of 3.2 m, which is 1.6 m.
02

Calculate Torques from Boys A and B

The torque for Boy A is calculated as \( \tau_A = m_A \times d_A = 45 \times 1.6 \). The torque for Boy B is \( \tau_B = m_B \times d_B = 35 \times 1.6 \). The torques are acting in opposite directions about the fulcrum.
03

Express Balance Condition

For the seesaw to balance, the total torque from the children must equal zero. So, the torque from Girl C must counterbalance the torque difference between Boy A and Boy B. Let \( d_C \) be the distance of Girl C from the fulcrum. Then, \( m_C \times d_C = \lvert \tau_A - \tau_B \rvert \).
04

Insert Values into Balance Equation

Substitute the known values into the balance equation: \( 25 \times d_C = \lvert 45 \times 1.6 - 35 \times 1.6 \rvert \). This simplifies to \( 25 \times d_C = \lvert 72 \rvert \).
05

Solve for Girl C's Position

Solve for \( d_C \): \( d_C = \frac{72}{25} \approx 2.88 \, \ ext{m} \). Thus, Girl C should place herself 2.88 m away from the fulcrum to balance the seesaw.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Torque Calculation
Torque is an essential concept when dealing with seesaws or any rotational systems. It is what causes an object to rotate around a pivot point, or fulcrum. To calculate torque, you need two main components:
  • The force applied, which in this case is the weight of the child (mass times gravitational acceleration).
  • The distance from the pivot point, where the force is applied.
The formula for torque is expressed as:\[ \tau = F \times d \]where \( \tau \) is the torque, \( F \) is the force applied, and \( d \) is the distance from the pivot point.In the seesaw example, the torques from the children are calculated by multiplying their masses (mass times gravity to get force) by their respective distances from the fulcrum. Boy A and Boy B contribute torques in opposite directions since they are on opposite ends of the seesaw. This balance of torques ensures that the seesaw can tip in either direction, depending on how these torques add up.
Finding the Center of Mass in a System
The center of mass is the point at which the mass of a system is balanced in all directions. It doesn't have to be located within the physical boundaries of the object, which is particularly important when considering a seesaw. A seesaw can be seen as a system of masses (two boys here), and the location of the center of mass will determine the balance point. Since both boys are of different masses, the center of mass will naturally shift towards the heavier boy, Boy A in this scenario. In a balanced seesaw, the fulcrum should ideally be placed at the center of mass of the combined system of all children involved. When Girl C joins, her position needs to alter in such a way that she moves the effective center of mass back to the seesaw’s original pivot point (the fulcrum), ensuring balanced equilibrium. Ultimately, understanding the center of mass helps us predict exactly where additional weight should be placed to achieve equilibrium.
Achieving Equilibrium Condition
Equilibrium occurs when all torques acting on the system cancel each other out, resulting in no net movement. This is a crucial condition for a seesaw to balance properly. For the seesaw to maintain equilibrium, the sum of clockwise torques must equal the sum of counterclockwise torques. This is the equilibrium condition for rotational systems. Mathematically, it can be expressed as:\[ \tau_{clockwise} = \tau_{counterclockwise} \]In the given exercise, this involves balancing the torque produced by Boy A on one end with the combined torques of Boy B and Girl C on the opposite side.When Girl C positions herself such that her torque balances the difference between torques from Boy A and Boy B, the seesaw achieves the overall equilibrium condition. This means that the rotational forces are balanced, and the seesaw itself will remain level without tipping to either side.

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Most popular questions from this chapter

Parachutists whose chutes have failed to open have been known to survive if they land in deep snow. Assume that a 75-kg parachutist hits the ground with an area of impact of \(0.30 m^2\) at a velocity of 55 m/s, and that the ultimate strength of body tissue is \(5 \times 10^5 N/m^2\). Assume that the person is brought to rest in 1.0 m of snow. Show that the person may escape serious injury.

A cube of side \(\ell\) rests on a rough floor. It is subjected to a steady horizontal pull \(F\), exerted a distance \(h\) above the floor as shown in Fig. 9-79. As \(F\) is increased, the block will either begin to slide, or begin to tip over. Determine the coefficient of static friction \(\mu_s\) so that (a) the block begins to slide rather than tip; (b) the block begins to tip. [\(Hint\): Where will the normal force on the block act if it tips?]

(III) A scallop forces open its shell with an elastic material called abductin, whose Young's modulus is about \(2.0 \times 10^6 N/m^2\). If this piece of abductin is 3.0 mm thick and has a cross-sectional area of \(0.50 cm^2\), how much potential energy does it store when compressed 1.0 mm?

(II) A 75-kg adult sits at one end of a 9.0-m-long board. His 25-kg child sits on the other end. \((a)\) Where should the pivot be placed so that the board is balanced, ignoring the board's mass? \((b)\) Find the pivot point if the board is uniform and has a mass of 15 kg.

(II) A 15-cm-long tendon was found to stretch 3.7 mm by a force of 13.4 N. The tendon was approximately round with an average diameter of 8.5 mm. Calculate Young's modulus of this tendon.

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