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Chapter 4: Problem 15

A truck driving on a level highway is acted on by the following forces: a downward gravitational force of \(52 \mathrm{kN}\) (kilonewtons); an upward contact force due to the road of \(52 \mathrm{kN} ;\) another contact force due to the road of \(7 \mathrm{kN},\) directed east; and a drag force due to air resistance of \(5 \mathrm{kN}\), directed west. What is the net force acting on the truck?

Short Answer

Expert verified
Answer: The net force acting on the truck is 2 kN in the eastward direction.

Step by step solution

01

Identify Horizontal and Vertical Components

To sum the forces vectorially, we need to split the four forces into horizontal and vertical components. Force 1: - Vertical component: \(52 \mathrm{kN}\) (downward) Force 2: - Vertical component: \(52 \mathrm{kN}\) (upward) Force 3: - Horizontal component: \(7 \mathrm{kN}\) (east) Force 4: - Horizontal component: \(5 \mathrm{kN}\) (west)
02

Calculate Net Vertical Force

To find the net vertical force (\(F_{v}\)), sum the vertical components of all four forces: \(F_{v} = -52 \mathrm{kN} + 52 \mathrm{kN} = 0 \mathrm{kN}\)
03

Calculate Net Horizontal Force

To find the net horizontal force (\(F_{h}\)), sum the horizontal components of all four forces: \(F_{h} = 7 \mathrm{kN} - 5 \mathrm{kN} = 2 \mathrm{kN}\)
04

Calculate the Net Force Acting on the Truck

Now we have the net horizontal force and the net vertical force, so we can calculate the net force (\(F_{net}\)) as follows: \(F_{net} = \sqrt{F_{v}^2 + F_{h}^2} = \sqrt{(0 \mathrm{kN})^2 + (2 \mathrm{kN})^2} = 2 \mathrm{kN}\) (east) The net force acting on the truck is \(2 \mathrm{kN}\) in the eastward direction.

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Most popular questions from this chapter

A box full of books rests on a wooden floor. The normal force the floor exerts on the box is \(250 \mathrm{N}\). (a) You push horizontally on the box with a force of \(120 \mathrm{N},\) but it refuses to budge. What can you say about the coefficient of static friction between the box and the floor? (b) If you must push horizontally on the box with a force of at least \(150 \mathrm{N}\) to start it sliding, what is the coefficient of static friction? (c) Once the box is sliding, you only have to push with a force of \(120 \mathrm{N}\) to keep it sliding. What is the coefficient of kinetic friction?
Felipe is going for a physical before joining the swim team. He is concerned about his weight, so he carries his scale into the elevator to check his weight while heading to the doctor's office on the 21 st floor of the building. If his scale reads \(750 \mathrm{N}\) while the elevator has an upward acceleration of \(2.0 \mathrm{m} / \mathrm{s}^{2},\) what does the nurse measure his weight to be?
A skier with a mass of \(63 \mathrm{kg}\) starts from rest and skis down an icy (frictionless) slope that has a length of \(50 \mathrm{m}\) at an angle of \(32^{\circ}\) with respect to the horizontal. At the bottom of the slope, the path levels out and becomes horizontal, the snow becomes less icy, and the skier begins to slow down, coming to rest in a distance of \(140 \mathrm{m}\) along the horizontal path. (a) What is the speed of the skier at the bottom of the slope? (b) What is the coefficient of kinetic friction between the skier and the horizontal surface?
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While an elevator of mass 832 kg moves downward, the tension in the supporting cable is a constant \(7730 \mathrm{N}\) Between \(t=0\) and \(t=4.00 \mathrm{s},\) the elevator's displacement is \(5.00 \mathrm{m}\) downward. What is the elevator's speed at \(t=4.00 \mathrm{s} ?\)
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