Problem 48 - An Introduction to the Mechanics of Solids in SI Units

Question

In building an orbiting space laboratory it will be necessary to drill holes in the flat steel wall of a space vehicle. The astronaut doing the drilling is unable to apply any appreciable amount either of force or of torque during the drilling so that it is necessary to mount the drill in a holder with three legs terminating in magnets, which grip the wall of the space vehicle. If the drilling torque is \(15 \mathrm{~N} \cdot \mathrm{m}\), and the normal force is \(50 \mathrm{~N}\), compute the minimum allowable holding force at each leg if the friction coefficient between the legs and space vehicle is \(0.4\).

Step-by-Step Solution

Answer

The minimum allowable holding force at each leg to prevent the drill from rotating is approximately \(6.67 \mathrm{N}\).

Step 1: Understand the Requirements

The problem requires the calculation of the minimum allowable holding force at each leg. The drill holder has three legs. The frictional force (\(F_f\)) created by each magnet must be enough to counter the drilling torque. The frictional force can be determined by the equation \(F_f=\mu \cdot F_n\), where \(\mu\) is the friction coefficient and \(F_n\) the normal force. The torque is calculated by \(T=F \cdot r\) where \(F\) is force and \(r\) is the distance to the point of rotation. However, we don't know the distance to the point of rotation and can't calculate the force (F). But the overall torque is equivalent to the sum of the torques created by each leg.

Step 2: Calculate the overall frictional force

The overall frictional force can be calculated by multiplying the normal force with the friction coefficient. This force is also equal to the drilling torque. Let's calculate:\[F_{\text{frictional overall}} = \mu \cdot F_n = 0.4 \cdot 50 N = 20N \]

Step 3: Calculate the minimum force per leg

The drill is hold by three legs, thus the force counteracting the drilling torque is equally distributed to the three legs. To find the minimum allowable holding force at each leg, we have to divide the overall frictional force by the number of legs. Let's calculate:\[F_{\text{frictional per leg}} = \frac{F_{\text{frictional overall}}}{\text{Number of Legs}} = \frac{20N}{3} \approx 6.67N \]

Key Concepts

Frictional Force Calculation

Understanding how to calculate frictional force is essential when dealing with the mechanics of solids. In the context of an astronaut drilling into a space vehicle's wall, the concept becomes critical to ensure the drill is securely held in place.

The frictional force, represented by the equation \(F_f = \text{mu} \cdot F_n\), is the product of the normal force (\(F_n\)) and the friction coefficient (\(\text{mu}\)). The friction coefficient is a dimensionless value that represents the amount of friction between two surfaces. It varies depending on the materials in contact. In our space drilling example, a coefficient of 0.4 implies a moderately high frictional force due in part to the magnetic attraction between the drill's legs and the steel wall.

The normal force in this scenario is provided by the magnets attached to each leg of the drill holder. Since the normal force is given as 50 N, and all legs are expected to exert an equal contribution to counteract the drilling torque, it's simply a matter of multiplying this normal force by the friction coefficient to find the overall frictional force each leg must achieve.

Drilling Torque in Space

The concept of torque is pivotal in mechanics and is especially interesting when applied to a scenario like drilling in space. Torque represents a rotational force, and its calculation is essential for understanding how to apply forces correctly in mechanical operations.

For calculating the torque induced by drilling, the equation usually used is \(T = F \cdot r\), where \(T\) is torque, \(F\) is the force applied, and \(r\) is the perpendicular distance from the axis of rotation to the point of force application. However, in our space lab problem, the astronaut cannot apply a force perpendicular to the point of rotation; consequently, the drill holder with magnetic legs is designed to produce the required torque to facilitate drilling.

Given the drilling torque is 15 Nm and the required frictional force to hold the drill stable is 20 N (as calculated previously), one can see that the magnets serve a dual purpose: exerting a normal force and generating frictional force to resist the rotational forces of the drill—allowing work to be done without the need for the astronaut to apply physical torque.

Normal Force and Friction Coefficient

To complete the picture, let's further explore the relationship between normal force and the friction coefficient. The normal force is the force exerted by a surface perpendicular to an object. In our example, it is the attractive magnetic force each leg exerts towards the steel wall.

The friction coefficient (\(\text{mu}\)), as we've seen, is a measure of how much frictional force is generated for a given normal force between two surfaces. A higher friction coefficient would result in a stronger frictional force, which could be beneficial in situations requiring a strong hold, like our astronaut's space drill. Conversely, a low friction coefficient would suggest less friction and, consequently, potentially inadequate grip for drilling tasks.

By the formula \(F_f = \text{mu} \cdot F_n\), which combines these two concepts, it provides a direct way to calculate the frictional force. This relationship allows us to determine the minimum holding force necessary at each leg, ensuring the drill does not slip or move during operation. Diving into these core concepts helps establish a strong foundation for problem-solving in the mechanics of solids, and reinforces the importance of friction and normal forces in everyday mechanical applications—whether on Earth or in the vastness of space.