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Chapter 11: Problem 6

Two people are carrying a uniform wooden board that is 3.00 \(\mathrm{m}\) long and weighs 160 \(\mathrm{N}\) . If one person applies an upward force equal to 60 \(\mathrm{N}\) at one end, at what point does the other person lift? Begin with a free-body diagram of the board.

Short Answer

Expert verified
The second person lifts 2.4 meters from the first person (where the 60 N force is applied).

Step by step solution

01

Understand the Problem and Draw the Free-body Diagram

We are given a uniform wooden board that is 3.00 meters long with a weight of 160 N. One person applies an upward force of 60 N at one end. We aim to find where the other person applies their force to keep the board in equilibrium. Start by drawing a free-body diagram, illustrating the forces involved: a 60 N force upwards at one end, a 160 N force (weight of the board) acting downwards at the center (1.5 m from either end), and an unknown upward force, \( F \), applied at a distance \( x \) from the opposite end.
02

Apply the Equilibrium Conditions

For the board to be in equilibrium, the sum of the vertical forces and the sum of the moments (torques) must be zero. The vertical force equilibrium condition can be expressed as: \(-160 \text{ N} + 60 \text{ N} + F = 0\), where \( F \) is the force applied by the second person.
03

Solve for the Vertical Force

From the equation in Step 2: \(-160 + 60 + F = 0\). Solving for \( F \), we find, \( F = 100 \text{ N}\).
04

Set Up the Torque Equilibrium Equation

Choose a pivot point to simplify calculations. Let's take the point where the 60 N force is applied as the pivot. Write the torque equation: \(-160 \text{ N} \times 1.5 \text{ m} + F \times x = 0\). The distance \( 1.5 \text{ m} \) is the distance from the pivot to the center of the board where the weight acts.
05

Solve for the Location \( x \) of the Second Force

Substitute \( F = 100 \text{ N} \) into the torque equation: \(-160 \times 1.5 + 100 \times x = 0\). Solving for \( x \), we get, \( x = \frac{160 \times 1.5}{100} = 2.4 \text{ m}\). This distance \( x \) is from the pivot where the 60 N force is applied, so the second person lifts the board 2.4 m from that end.

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Key Concepts

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

Free-body Diagram
A free-body diagram is a crucial tool in physics that helps visualize the forces acting on an object. It simplifies the problem by representing all the acting forces. In our exercise, the free-body diagram of the wooden board helps us identify where each force is applied and understand their directions.
  • Begin by drawing a simple rectangle to represent the board. The entire length is 3.00 meters.
  • Mark a point at the center (1.5 meters from either end) where the weight of 160 N acts downwards.
  • At one end of the board, draw an arrow pointing upwards to denote the 60 N force applied by one person.
  • Mark another point on the board where the second person's force, labeled as \( F \), acts upwards at an unknown distance \( x \) from the end opposite the 60 N force.
Understanding the placement of these forces through the free-body diagram is vital for applying equilibrium conditions and solving for the unknowns.
Equilibrium Conditions
Equilibrium is achieved when there is no net force or torque acting on an object. This means that the object remains at rest or moves with constant velocity. For our wooden board to be in perfect balance without rotation or movement, it must satisfy two conditions.
  • **Sum of Vertical Forces**: The total of all upward forces must equal the total of all downward forces. For our board: \(-160 \, \text{N} + 60 \, \text{N} + F = 0\).
  • **Sum of Torques**: The total torque acting around any pivot point must be zero. This ensures the object does not rotate. In this exercise, we took the point where 60 N force is applied as the pivot for simplicity.
By ensuring these conditions are met, we can solve for any unknowns, like the location \( x \) where the second person's force is applied.
Force Balance
The concept of force balance plays an essential role in solving physical problems involving equilibrium. Force balance means that the forces acting on an object are equal and opposite, ensuring the object remains stationary or in constant motion. In our exercise, this concept was used to determine how two forces balance the weight of the board.
  • To achieve balance, the upward forces (applied by the two people) must sum up to equal the downward force (the weight of the board).
  • The equation \(-160 + 60 + F = 0\) enabled us to solve for the unknown force \( F \), giving us \( F = 100 \, \text{N}\).
Force balance is integral to ensuring the board does not tip over or move from its resting position, making it an essential part of understanding torque equilibrium scenarios.

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Most popular questions from this chapter

BIO Forearm. In the human arm, the forearm and hand pivot about the elbow joint. Consider a simplified model in which the biceps muscle is attached to the forearm 3.80 \(\mathrm{cm}\) from the elbow joint. Assume that the person's hand and forearm together weigh 15.0 \(\mathrm{N}\) and that their center of gravity is 15.0 \(\mathrm{cm}\) from the elbow (not quite halfway to the hand). The forearm is held horizontally at a right angle to the upper arm, with the biceps muscle exerting its force perpendicular to the forearm. (a) Draw a free-body diagram for the forearm, and find the force exerted by the biceps when the hand is empty. (b) Now the person holds a 80.0 -N weight in his hand, with the forearm still horizontal. Assume that the center of gravity of this weight is 33.0 \(\mathrm{cm}\) from the elbow. Construct a free-body diagram for the forearm, and find the force now exerted by the biceps. Explain why the biceps muscle needs to be very strong. (c) Under the conditions of part (b), find the magnitude and direction of the force that the elbow joint exerts on the forearm. (d) While holding the \(80.0-\mathrm{N}\) weight, the person raises his forearm until it is at an angle of \(53.0^{\circ}\) above the horizontal. If the biceps muscle continues to exert its force perpendicular to the forearm, what is this force when the forearm is in this position? Has the force increased or decreased from its value in part (b)? Explain why this is so, and test your answer by actually doing this with your own arm.

A solid gold bar is pulled up from the hold of the sunken RMS Titanic. (a) What happens to its volume as it goes from the pressure at the ship to the lower pressure at the ocean's surface? (b) The pressure difference is proportional to the depth. How many times greater would the volume change have been had the ship been twice as deep? (c) The bulk modulus of lead is one- fourth that of gold. Find the ratio of the volume change of a solid lead bar to that of a gold bar of equal volume for the same pressure change.

Suppose that you can lift no more than 650 \(\mathrm{N}\) (around 150 lb) unaided. (a) How much can you lift using a \(1.4-\mathrm{m}\) -long wheelbarrow that weighs 80.0 \(\mathrm{N}\) and whose center of gravity is 0.50 \(\mathrm{m}\) from the center of the wheel (Fig. E11.16)? The center of gravity of the load carried in the wheelbarrow is also 0.50 \(\mathrm{m}\) from the center of the wheel. (b) Where does the force come from to enable you to lift more than 650 \(\mathrm{N}\) using the wheelbarrow?

If you put a uniform block at the edge of a table, the center of the block must be over the table for the block not to fall off. (a) If you stack two identical blocks at the table edge, the center of the top block must be over the bottom block, and the center of gravity of the two blocks together must be over the table. In terms of the length \(L\) of each block, what is the maximum overhang possible (Fig. \(P 11.78 ) ?\) (b) Repeat part (a) for three identical blocks and for four identical blocks. (c) Is it possible to make a stack of blocks such the uppermost block is not directly over the table at all? How many blocks would it take to do this? (Try this with your friends using copies of this book.)

Bulk Modulus of an Ideal Gas. The equation of state (the equation relating pressure, volume, and temperature) for an ideal gas is \(p V=n R T,\) where \(n\) and \(R\) are constants. (a) Show that if the gas is compressed while the temperature \(T\) is held constant, the bulk modulus is equal to the pressure. (b) When an ideal gas is compressed without the transfer of any heat into or out of it, the pressure and volume are related by \(p V^{\gamma}=\) constant, where \(\gamma\) is a constant having different values for different gases. Show that, in this case, the bulk modulus is given by \(B=\gamma p\)
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