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Chapter 11: Problem 16

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?

Short Answer

Expert verified
(a) You can lift 1090 N using the wheelbarrow. (b) The wheelbarrow provides mechanical advantage.

Step by step solution

01

Understanding the problem

You need to determine the maximum load you can lift using a wheelbarrow given the constraints. This involves using the principle of moments (torque) around the pivot (the wheel). With a maximum lifting force of 650 N, you need to find where this force is exerted and calculate the additional force you can handle using the wheelbarrow's leverage.
02

Define known values and leverage

The wheelbarrow's length is 1.4 m, and your maximum lifting force (F) is 650 N. Weights of the wheelbarrow and the load act as additional forces: 80 N and a load weight (W) we are solving for. Both the wheelbarrow's and load's centers of gravity are 0.5 m from the wheel.
03

Set up the torque equation

Torque (moment) is given as force times distance (lever arm). The wheel acts as the pivot point. Calculate moments around the pivot:- For the load: Moment = W * 0.5 m.- For the wheelbarrow: Moment = 80 N * 0.5 m.- Your lifting force: Moment = 650 N * (1.4 m - 0.5 m). Set the sum of these torques to zero for the wheelbarrow to be in rotational equilibrium: \[ W \times 0.5 + 80 \times 0.5 = 650 \times 0.9 \].
04

Solve the torque equation

Substitute known values into the torque equilibrium equation and solve for W: \[ W \times 0.5 + 80 \times 0.5 = 650 \times 0.9 \]. Simplifies to: \[ 0.5W + 40 = 585 \] Subtract 40 from both sides: \[ 0.5W = 545 \]. Divide by 0.5: \[ W = 1090 \; \text{N} \].
05

Source of additional lifting force

The additional force you can lift comes from the mechanical advantage provided by the wheelbarrow. The wheelbarrow allows you to exert more force by using the longer lever formed between the load and the point where you apply force.

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Key Concepts

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

Understanding Torque
Torque is an important concept in physics that deals with rotational forces. It essentially captures how much a force causes an object to rotate. Imagine opening a door: pushing at the edge makes it swing easily compared to pushing near the hinges.
To calculate torque, you multiply the force applied by the distance from the pivot point, known as the lever arm. The formula is given by \[ \text{Torque} = \text{Force} \times \text{Lever Arm}\]In scenarios like using a wheelbarrow, torque plays a crucial role. The pivot point would be the wheel, and understanding torque helps us use the wheelbarrow effectively to lift heavier loads than possible by hand alone.
Finding Rotational Equilibrium
Rotational equilibrium occurs when all the torques acting on an object cancel each other out, leading to no net rotation. This is vital for ensuring the wheelbarrow doesn't tip over when loaded.
To recall, when in rotational equilibrium, the sum of clockwise torques equals the sum of counterclockwise torques around the pivot. For our wheelbarrow example:
  • Your lifting torque acts in one direction, and must be balanced by the torque of the load and the weight of the wheelbarrow acting in the opposite direction.
  • Mathematically, this means:\[W \times \text{load lever arm} + \text{wheelbarrow weight} \times \text{its lever arm} = \text{lifting force} \times \text{its lever arm}\]
  • By setting these forces to balance, you ensure the system remains stable without rotation.
Lever Arm Mechanics
The concept of the lever arm is pivotal when dealing with torque and rotational motion. The lever arm is the distance from the pivot point to the point where the force is applied.
In mechanical systems like wheelbarrows, the lever arm determines how effectively force is applied. The longer the lever arm, the more torque you can generate with the same amount of force. Consider:
  • When using a wheelbarrow, the distance from the wheel (pivot) to where you hold and lift forms your lever arm.
  • In our problem setup, having a lever arm of 0.9 m means your lifting force is more effective, making it easier to lift heavy loads with less effort.
This is why extending the handle of a wheelbarrow can make a noticeable difference in handling heavy contents.
The Physics of Wheelbarrows
Wheelbarrows are brilliant examples of how simple machines make heavy tasks easier. They operate on principles of leverage and rotational equilibrium to give users a mechanical advantage.
Here's how they function:
  • The wheel acts as a fulcrum, and the handles as levers; thus, force applied at the handles is magnified to lift loads in the tray.
  • By distributing the weight over the wheel, wheelbarrows reduce the effort required, compared to directly lifting the load.
  • This mechanical advantage means that even heavy loads can be moved with relatively little force, as long as balance (rotational equilibrium) is maintained.
By understanding these mechanics, you can use wheelbarrows more effectively, reducing physical strain while increasing productivity.

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Most popular questions from this chapter

BIO Leg Raises. In a simplified version of the musculature action in leg raises, the abdominal muscles pull on the femur (thigh bone) to raise the leg by pivoting it about one end (Fig. P11.57. When you are lying. horizontally, these muscles make an angle of approximately \(5^{\circ}\) with the femur, and if you raise your legs, the muscles remain approximately horizontal, so the angle \(\theta\) increases. We shall assume for simplicity that these muscles attach to the femur in only one place, 10 \(\mathrm{cm}\) from the hip joint (although, in reality, the situation is more complicated). For a certain \(80-\mathrm{kg}\) person having a leg 90 \(\mathrm{cm}\) long, the mass of the leg is 15 \(\mathrm{kg}\) and its center of mass is 44 \(\mathrm{cm}\) from his hip joint as measured along the leg. If the person raises his leg to \(60^{\circ}\) above the horizontal, the angle between the abdominal muscles and his femur would also be about \(60^{\circ} .\) (a) With his leg raised to \(60^{\circ},\) find the tension in the abdominal muscle on each leg. As usual, begin your solution with a free-body diagram. (b) When is the tension in this muscle greater: when the leg is raised to \(60^{\circ}\) or when the person just starts to raise it off the ground? Why? (Try this yourself to check your answer.) (c) If the abdominal muscles attached to the femur were perfectly horizontal when a person was lying down, could the person raise his leg? Why or why not?

BIO 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 length 0.200 \(\mathrm{m}\) and cross-sectional area 50.0 \(\mathrm{cm}^{2} .\)

Two people carry a heavy electric motor by placing it on a light board 2.00 \(\mathrm{m}\) long. One person lifts at one end with a force of \(400 \mathrm{N},\) and the other lifts the opposite end with a force of 600 \(\mathrm{N}\) . (a) What is the weight of the motor, and where along the board is its center of gravity located? (b) Suppose the board is not light but weighs \(200 \mathrm{N},\) with its center of gravity at its center, and the two people each exert the same forces as before. What is the weight of the motor in this case, and where is its center of gravity located?

A uniform 300 -N trapdoor in a floor is hinged at one side. Find the net upward force needed to begin to open it and the total force exerted on the door by the hinges (a) if the upward force is applied at the center and (b) if the upward force is applied at the center of the edge opposite the hinges.

A uniform, \(90.0-\mathrm{N}\) table is 3.6 \(\mathrm{m}\) long, 1.0 \(\mathrm{m}\) high, and 1.2 \(\mathrm{m}\) wide. \(\mathrm{A} 1500\) -N weight is placed 0.50 \(\mathrm{m}\) from one end of the table, a distance of 0.60 \(\mathrm{m}\) from each side of the table. Draw a free body diagram for the table and find the force that each of the four legs exerts on the floor.
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