12.5: Chapter 12 Homework Problems
Exercise \(\PageIndex{1}\)
The SUV shown below has an initial velocity of 90 ft/s. It slams on its brakes, coming to a stop over a distance of 300 feet. If the car has a weight of 3500 lbs and a center of mass as shown below, what are the normal forces at the front wheels? What are the normal forces at the back wheels?
- Solution
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\( F_{N_{rear}} = 1291.4 \ lbs\)
\( F_{N_{front}} = 2208.6 \ lbs\)
Exercise \(\PageIndex{2}\)
A ring-shaped space station can be approximated as a thin ring 60 meters in diameter with a mass of 500,000 kg. The space station has a set of thrusters able to exert equal and opposite forces as shown below. If we want to cause an angular acceleration of 0.1 rad/s² in the space station, what is the force required from each thruster?
- Solution
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\(F_{thruster} = 750 \ kN\)
Exercise \(\PageIndex{3}\)
A 50-kg barrel with a diameter of 0.75 meters is placed on a 20-degree slope. Assuming the barrel rolls without slipping, what will the acceleration of the barrel's center of mass be?
- Solution
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\(a_x = 2.24 \ m/s^2\)
Exercise \(\PageIndex{4}\)
A 3-meter-long, 25-kg beam is supported by two cables as shown below. You can treat the beam as a slender rod. Assume that we want the left end of the beam at point A to remain at a constant height while the right end of the beam at point B accelerates upwards at a rate of 1 m/s².
- What is the rate of acceleration of the center of the beam and the rate of angular acceleration for the beam?
- What will \(T_1\) and \(T_2\) need to be to achieve these accelerations?
- Solution
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\( a_{C\y} = 0.5 \ m/s^2, \, \alpha = 0.333 \ \frac{rad}{s} \)
\( T_1 = 81.75 \ N, \, T_2 = 176 \ N \)
Exercise \(\PageIndex{5}\)
You are modeling the robotic arm shown below. Treat each section of the arm as a slender rod. Section OA weighs 30 lbs and section AB weighs 18 lbs. If we want the relative angular accelerations and velocities shown below, what should the motor torques be at O and A? (This is a top-down view of the robot arm.)
- Solution
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\(M_O = - 3.9 \ ft\)-\(lbs\)
\(M_A = -19.3 \ ft\)-\(lbs\)