Question

A flatbed truck is loaded with a stack of sacks of cement whose combined mass is \(1143.5 \mathrm{~kg}\). The coefficient of static friction between the bed of the truck and the bottom sack in the stack is \(0.372,\) and the sacks are not tied down but held in place by the force of friction between the bed and the bottom sack. The truck accelerates uniformly from rest to \(56.6 \mathrm{mph}\) in \(22.9 \mathrm{~s}\). The stack of sacks is \(1 \mathrm{~m}\) from the end of the truck bed. Does the stack slide on the truck bed? The coefficient of kinetic friction between the bottom sack and the truck bed is \(0.257 .\) What is the work done on the stack by the force of friction between the stack and the bed of the truck?

Short answer

(a) 1.10 m/s² 2. What is the maximum static friction force between the stack and the truck bed? (b) 4123 N 3. Does the stack of cement sacks slide on the truck bed? (c) No 4. What is the work done on the stack by the force of friction between the stack and the truck bed? (d) 1258 J

Step-by-step solution

Calculate the acceleration of the truck

First, we need to convert the truck's final velocity from mph to m/s using the conversion factor: \(1\mathrm{~mph} = 0.44704\mathrm{~m/s}\). Thus, \(56.6\mathrm{~mph} \times 0.44704\mathrm{~m/s/mph} \approx 25.3\mathrm{~m/s}\). Now, we can use the formula for uniformly accelerated motion: \(v = u + at\), where \(v\) is the final velocity, \(u\) is the initial velocity (which is zero since the truck starts from rest), \(a\) is the acceleration, and \(t\) is the time taken. Solving for acceleration, \(a = (v - u) / t\), we get: \(a = (25.3 - 0) / 22.9 \approx 1.10\mathrm{~m/s^2}\).

Calculate the maximum static friction force

Next, we will find the maximum static friction force between the bottom sack and the truck bed using the formula: \(f_s = \mu_s m g\), where \(\mu_s\) is the coefficient of static friction, \(m\) is the combined mass of the sacks, and \(g\) is the acceleration due to gravity (\(9.81\mathrm{~m/s^2}\)). \(f_s = 0.372 \times 1143.5 \times 9.81 \approx 4123\mathrm{~N}\).

Calculate the force required to prevent sliding

Now, we need to find the force required to prevent sliding using Newton's second law: \(F = ma\), where \(F\) is the force, and \(a\) is the acceleration calculated in step 1. \(F = 1143.5 \times 1.10 \approx 1258\mathrm{~N}\).

Determine if the stack slides

We compare the maximum static friction force (\(f_s\)) and the force required to prevent sliding (\(F\)). If \(F > f_s\), the stack will slide. Otherwise, the stack will not slide. Since \(1258\mathrm{~N}\) is less than \(4123\mathrm{~N}\), the stack does not slide. Now, let's find the work done on the stack by the force of friction between the stack and the truck bed.

Calculate the work done by the frictional force

Since the sacks do not slide, static friction force is acting between the truck bed and the stack. Hence, the friction force acting on the stack is equal to the force required to prevent sliding, which is 1258 N. The work done by the frictional force can be calculated by using the formula: \(W = Fd\cos{\theta}\), where \(W\) is the work done, \(F\) is the force, \(d\) is the distance, and \(\theta\) is the angle between the force and displacement directions. Since the force and displacement are in the same direction, \(\theta = 0°\). Therefore, \(W = 1258 \times 1 \times \cos{0°} = 1258\mathrm{~J}\). In conclusion, the stack of cement sacks does not slide on the truck bed, and the work done on the stack by the force of friction between the stack and the bed of the truck is 1258 J.

Key concepts

Coefficient of Static Friction

Understanding the coefficient of static friction is crucial when studying how objects interact with surfaces. It is a dimensionless number that represents the ratio of the maximum static frictional force between two surfaces to the normal force pressing them together. In mathematical terms, the coefficient of static friction, denoted by \( \mu_s \), can be expressed as \( f_s = \mu_s N \), where \( f_s \) is the static frictional force and \( N \) is the normal force.

In our example with the cement sacks on the truck, \( \mu_s \) was given as 0.372. We calculated the static frictional force using the mass of the sacks and the acceleration due to gravity to determine if the sacks would slide when the truck accelerated. If the calculated frictional force is greater than the force needed to move the sacks, they will not slide; this was indeed the case, reinforcing the importance of \( \mu_s \) in predicting motion between surfaces.

Newton's Second Law

Newton's second law of motion is a fundamental principle which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass \( (a = F/m) \). In simpler terms, the more force you apply to an object, the faster it will go, if you disregard its mass.

Applying this law to our truck and cement sack scenario, we used the calculated acceleration and the mass of the sacks to determine the force needed to prevent sliding. This force then allowed us to determine whether the static friction was enough to stop the sacks from sliding when the truck accelerated. The successful application of Newton's second law provided a clear quantitative measure influencing the outcome of the scenario.

Work Done by Friction

Work done by friction is a concept that describes the energy transfer when a force is applied over a distance, with friction opposing the motion. The formula for work done is \( W = Fd\cos{\theta} \), wherein \( W \) is the work, \( F \) is the force, \( d \) is the distance, and \( \theta \) is the angle between the force applied and the direction of motion. When the direction of friction and the movement are the same, \( \theta = 0° \), simplifying the work done to \( W = Fd \).

In the case of the cement sacks on the truck bed, we calculated the work done by the static friction force to be 1258 Joules. The direction of the static friction was the same as the truck's motion; hence no kinetic energy was lost due to sliding, and the work done could be calculated using straightforward multiplication.

Uniformly Accelerated Motion

Uniformly accelerated motion refers to motion in which an object experiences a constant acceleration. This typically results in a change in velocity over time, which can be described by the formula \( v = u + at \). In this equation, \( v \) is the final velocity, \( u \) is the initial velocity, \( a \) is the acceleration, and \( t \) is the time elapsed.

For the truck's motion, we considered it to be uniformly accelerating from rest (initial velocity of zero) to a certain speed over a specific time period. By utilizing the concept of uniformly accelerated motion, we were able to calculate the truck's acceleration, which was then applied within Newton's second law and other relevant calculations to complete our analysis of the truck's cargo transport scenario.