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Chapter 15: Problem 133

The tractor together with the empty tank has a total mass of \(4 \mathrm{Mg}\). The tank is filled with \(2 \mathrm{Mg}\) of water. The water is discharged at a constant rate of \(50 \mathrm{~kg} / \mathrm{s}\) with a constant velocity of \(5 \mathrm{~m} / \mathrm{s}\), measured relative to the tractor. If the tractor starts from rest, and the rear wheels provide a resultant traction force of \(250 \mathrm{~N}\), determine the velocity and acceleration of the tractor at the instant the tank becomes empty.

Short Answer

Expert verified
The velocity of the tractor when the tank becomes empty is \(0 m/s\) and the acceleration is \(0 m/s虏\).

Step by step solution

01

Analyzing and gathering given data

The total mass of the tractor and the empty tank is \(4 Mg = 4000 kg\).The tank initially contains \(2 Mg = 2000 kg\) of water which is discharged at a rate of \(50 kg/s\) at a velocity of \(5 m/s\), and the driving force on the tractor is \(250 N\). What is needed is the velocity and acceleration of the tractor when the tank is empty.
02

Calculating the time taken to empty the tank

The total mass of water is \(2000 kg\) and it is discharging at the rate of \(50 kg/s\). So, the time taken to empty the tank can be computed as: \[ t = \frac{Mass_{water}}{rate_{discharge}} = \frac{2000}{50} = 40 s\]
03

Determine the reduction in mass per unit time

Mass of water is discharging at the rate of \(50 kg/s\). Therefore, decrease in mass of the system (rear wheels providing a resultant traction force) per unit time is \(50 kg/s\).
04

Applying Newton's Second Law of Motion

The net force in the system equals to the increase in momentum per unit time. In this case, it equals the sum of the reduction of mass per time and reduction in velocity of water per unit time. This force also equals to the mass of the system times the acceleration \(a\), as \(F = m \cdot a\). According to the problem, we know the net force equals \(250 N\). So, we can express the equation as follows: \[250 = (4000 - 50t) \cdot a + 50 \cdot 5 \] Where: \(m = 4000 - 50t\) is the mass of the system (tractor and the remaining water in the tank) at time \(t\), and \(50 \cdot 5 = 250\) is the force exerted by the discharging water (due to its velocity).
05

Determine the velocity and acceleration of the tractor

From the previous step, we derived a differential equation. Solving this differential equation gives us: \[a = \frac{250 - 250}{4000 - 50 \cdot 40} = 0 m/s虏\] When the tank gets empty (\(t = 40s\)), the acceleration of the tractor is \(0 m/s虏\). Then, to find the velocity of the tractor when the tank is empty, we integrate the acceleration over time (from 0 to 40 seconds), which gives: \[v = \int a dt = \int_0^{40} 0 dt = 0 m/s\] So, when the tank gets empty, the tractor stops (velocity equals \(0 m/s\)).

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Most popular questions from this chapter

A snowblower having a scoop \(S\) with a crosssectional area of \(A_{\mathrm{s}}=0.12 \mathrm{~m}^{3}\) is pushed into snow with a speed of \(v_{s}=0.5 \mathrm{~m} / \mathrm{s}\). The machine discharges the snow through a tube \(T\) that has a cross-sectional area of \(A_{T}=0.03 \mathrm{~m}^{2}\) and is directed \(60^{\circ}\) from the horizontal. If the density of snow is \(\rho_{s}=104 \mathrm{~kg} / \mathrm{m}^{3}\), determine the horizontal force \(P\) required to push the blower forward, and the resultant frictional force \(F\) of the wheels on the ground, necessary to prevent the blower from moving sideways. The wheels roll freely.

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