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(II) The bolts on the cylinder head of an engine require tightening to a torque of 95 m\(\cdot\)N. If a wrench is 28 cm long, what force perpendicular to the wrench must the mechanic exert at its end? If the six-sided bolt head is 15 mm across (Fig. 8-44), estimate the force applied near each of the six points by a wrench.

Short Answer

Expert verified
The mechanic must exert approximately 339.29 N perpendicularly on the wrench, with around 56.55 N at each of the six points of the bolt head.

Step by step solution

01

Understanding Torque

Torque is calculated as the product of the force applied and the perpendicular distance from the point of rotation. The formula is given by \[ \tau = F \times d \]where \( \tau \) is the torque, \( F \) is the force, and \( d \) is the distance from the pivot point (the length of the wrench in this case).
02

Convert Units

Convert the length of the wrench from centimeters to meters, since torque is given in m\(\cdot\)N. The wrench is 28 cm long, which equates to 0.28 meters.
03

Solve for Force

Rearrange the torque equation to solve for the force: \[ F = \frac{\tau}{d} \]Substitute the values, \( \tau = 95 \text{ m}\cdot\text{N} \) and \( d = 0.28 \text{ m} \), to find the force \[ F = \frac{95}{0.28} \approx 339.29 \text{ N} \].
04

Understanding the Bolt Head

The six-sided bolt head has a size across of 15 mm, which is not directly relevant to calculating the force needed on the wrench but is necessary to understand the fit and pressure distribution when using the wrench. We need each point of contact where the force will be distributed.
05

Calculate Force Per Point

Estimate the force applied near each of the six points by dividing the force by the number of points: \[ F_{point} = \frac{F}{6} = \frac{339.29}{6} \approx 56.55 \text{ N} \].

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Key Concepts

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

Force Distribution
When a force is applied to a bolted joint using a wrench or any tool, the distribution of that force is critical for ensuring stability and effectiveness. A wrench distributes the applied force over specific contact points, often aligned with the shape of the bolt head, such as a hexagon for a six-sided bolt.
The total force exerted by the mechanic is not applied uniformly across the entire surface of the bolt head. Instead, it's divided among the contact points where the wrench grips the bolt. This is crucial because if the force were not distributed evenly, it could lead to stripping or damaging the bolt head.
  • With a six-sided bolt head, the total force is split into six parts.
  • This force distribution helps ensure that no single point bears an excessive load.
  • Contact points spread out the force, reducing the risk of bolt or tool damage.
Proper understanding of force distribution allows for the correct application of torque, ensuring that mechanical components are secured without over-tightening or under-tightening.
Wrench Mechanics
Wrench mechanics involve the interaction between torque, force, and the distance from the pivot point (the length of the wrench). It's essential to know how these factors interplay, especially when tightening bolts to a specified torque. Let's delve into how these components work together.

When using a wrench, the goal is to apply a force that results in a specified torque. The wrench acts as a lever, with the bolt head being the pivot point. The longer the wrench, the greater the torque generated for the same amount of force applied.
  • Torque is the rotational equivalent of linear force. It measures how force causes an object to rotate.
  • Using a longer wrench reduces the force needed because torque increases with the increase in distance from the pivot point: \, \( \tau = F \times d \).
  • Wrench mechanics capitalize on leverage, enhancing the force applied at a greater distance.
This principle of using leverage to amplify force is foundational in mechanics and key to efficiently applying the correct torque when tightening or loosening bolts.
Unit Conversion
In many fields, including mechanics and physics, correct unit conversion is vital. This ensures that calculations are accurate and align with standard units, making it easier to communicate and understand results.

When calculating torque, it's common to work in mixed units — such as centimeters for length and Newton-meters for torque. However, calculations must use consistent units, requiring the conversion of length measurements into meters.
  • Always convert the wrench length from centimeters to meters by dividing by 100 — for example, 28 cm to 0.28 meters.
  • Unit conversion minimizes errors by maintaining consistency and standardization in formulas, ensuring accuracy.
  • Helps in comparing results and applying them across different scenarios or combining with other calculations.
By converting all measurements to standardized units, mechanics can be more precise, leading to better outcomes when applying forces such as torque.

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Most popular questions from this chapter

(III) A hammer thrower accelerates the hammer \((mass = 7.30 kg)\) from rest within four full turns (revolutions) and releases it at a speed of 26.5 m/s. Assuming a uniform rate of increase in angular velocity and a horizontal circular path of radius 1.20 m, calculate \((a)\) the angular acceleration, \((b)\) the (linear) tangential acceleration, \((c)\) the centripetal acceleration just before release, \((d)\) the net force being exerted on the hammer by the athlete just before release, and \((e)\) the angle of this force with respect to the radius of the circular motion. Ignore gravity.

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