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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 the board at 0.375 m from the opposite end.

Step by step solution

01

Visualize the Problem with a Free-body Diagram

Before solving the problem, sketch a free-body diagram. Draw the wooden board horizontally and label the forces acting on it. Label the weight of the board (160 N), acting downward at its center due to its uniformity, and the upward force of 60 N applied by the first person at one end of the board. Let the distance from the other end where the second person applies their force upwards be indicated as \(x\).
02

Understanding the Concept of Moments

Recall that the board is in static equilibrium; therefore, the sum of all forces and moments (torques) acting on it must be zero. The moment around any point on the board should sum to zero. Let's take moments around the point where the second person applies force to simplify the calculation.
03

Calculate Sum of Moments Around the Second Person

Use the principle of moments (torque) about the point where the second person applies force. Let \(F_2\) be the upward force exerted by the second person (unknown for now). The equation is:\[F_1 \times 3.00 - 160N \times (1.50 - x) = 0\]Where \(F_1 = 60 \text{ N}\) and \(1.50 \text{ m}\) is the center of the board.
04

Solve for \(x\)

Since the board is 3.00 m long and uniform, the center of mass is at its midpoint (1.50 m). For equilibrium of moments about the second person鈥檚 point:\[60 \text{ N} \times 3.00 \text{ m} - 160 \text{ N} \times (1.50 \text{ m} - x) = 0\]Which simplifies to:\[180 = 160 \times (1.50 - x)\]Solve for \(x\):\[x = 1.50 - \frac{180}{160}\]Simplify:\[x = 1.50 - 1.125 = 0.375 \text{ m}\]
05

Verify Total Force Cancellation

Ensure that the total vertical forces are zero to further verify equilibrium. The second person applies an upward force \(F_2 = 160 \text{ N} - 60 \text{ N} = 100 \text{ N}\). Sum of upward forces should equal the downward force due to gravity on the board.

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Key Concepts

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

Free-body Diagram
When analyzing problems in statics, the free-body diagram (FBD) is an essential tool. It helps to visualize all the forces acting upon an object. In our exercise, the object is a 3-meter long uniform wooden board. Start by sketching the board horizontally. Next, label the forces acting on this board.

These forces include:
  • The weight of the board, 160 N, acting downward at the center of the board because it is uniform.
  • An upward force of 60 N at one end, applied by the first person.
  • The force applied by the second person, acting upwards at some distance, denoted as \(x\) from the end the first person lifts.
Visualizing these forces with a free-body diagram forms the foundation for further analysis. It ensures all potential reactions and applied forces are considered before moving to calculation phases.
Moment of Force
The moment of force, or torque, is crucial for understanding balance and rotation in structures. In this exercise, you must grasp how moments influence equilibrium. A moment is created when a force causes an object to rotate around a point. The moment of force is calculated by multiplying the force by the perpendicular distance from the line of action of the force to the point of rotation, represented mathematically as \(Moment = Force \times Distance\).

As the board is in equilibrium, the sum of all moments around any point is zero. We simplify the calculations by taking moments around the point where the second person applies their force. This approach allows us to more directly solve for the unknown distance \(x\). It's essential to balance these moments because any imbalance would cause rotation, disrupting the board's stability.
Equilibrium of Moments
For the board to remain in static equilibrium, the sum of the moments of all forces about any point must equal zero. In this problem, considering the point where the second person applies her force helps simplify the mathematics.

Here鈥檚 the mathematical approach to ensuring equilibrium:
  • The total moment generated by the upward force of 60 N is given by \(60 \text{ N} \times 3.00 \text{ m}\).
  • This must balance the moment caused by the weight of the board, \(160 \text{ N} \times (1.50 \text{ m} - x)\), where 1.50 meters is the distance from the pivot to the board's center of gravity.
By solving the equilibrium equation, you find that the second person applies her lift at 0.375 meters from the end opposing the first person. This balance of moments is the crux of maintaining equilibrium, ensuring the board does not rotate or tip.

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Most popular questions from this chapter

Prior to being placed in its hole, a \(5700-\mathrm{N}, 9.0\) -m-long, unjform utility pole makes some nonzero angle with the vertical. A vertical cable attached 2.0 \(\mathrm{m}\) below its upperend holds it in place while its lower end rests on the ground. (a) Find the tension in the cable and the magnitude and direction of the force exerted by the ground on the pole. (b) Why don't we need to know the angle the pole makes with the vertical, as long as it is not zero?

A uniform drawbridge must be held at a \(37^{\circ}\) angle above the horizontal to allow ships to pass underneath. The drawbridge weighs \(45,000 \mathrm{N}\) and is 14.0 \(\mathrm{m}\) long. A cable is connected 3.5 \(\mathrm{m}\) from the hinge where the bridge pivots (measured along the bridge) and pulls horizontally on the bridge to hold it in place. (a) What is the tension in the cable? (b) Find the magnitude and direction of the force the hinge exerts on the bridge.

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 \(B=\gamma p\) .

A \(15,000-N\) crane pivots around a friction-free axle at its base and is supported by a cable making a \(25^{\circ}\) angle with the crane (Fig. 11.29 ). The crane is 16 \(\mathrm{m}\) long and is not uniform, its center of gravity being 7.0 \(\mathrm{m}\) from the axle as measured along the crane. The cable is attached 3.0 \(\mathrm{m}\) from the upper end of the crane. When the crane is raised to \(55^{\circ}\) above the horizontal holding an \(11,000-N\) pallet of bricks by a 2.2 -in very light cord, find (a) the tension in the cable, and \((b)\) the horizontal and vertical components of the force that the axle exerts on the crane. Start with a free-body diagram of the crane.

Biceps Muscle. A relaxed biceps muscle requires a force of 25.0 \(\mathrm{N}\) for an elongation of 3.0 \(\mathrm{cm}\) ; the same muscle under maximum tension requires a force of 500 \(\mathrm{N}\) for the same elongation. Find Young's modulus for the muscle tissue under each of these conditions if the muscle is assumed to be a uniform cylinder with blength 0.200 \(\mathrm{m}\) m and cross- sectional area 50.0 \(\mathrm{cm}^{2} .\)
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