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Chapter 4: Problem 20

A man is lazily floating on an air mattress in a swimming pool. If the weight of the man and air mattress together is \(806 \mathrm{N},\) what is the upward force of the water acting on the mattress?

Short Answer

Expert verified
Answer: The upward force of the water acting on the man and the air mattress is 806 N.

Step by step solution

01

Identify the forces acting on the system

In this problem, we have two forces acting on the man and the air mattress: the downward force due to the weight of the system (gravity) and the upward force due to the water (buoyancy).
02

Apply the concept of equilibrium

Since the man and the air mattress are lazily floating, it means they are in equilibrium. In equilibrium, the downward force (weight) is equal to the upward force (buoyancy), so we can write this as: \(F_{upward} = F_{downward}\)
03

Use the given weight to determine the upward force

We know the downward force is the weight of the man and the air mattress, which is 806 N. Now we can equate it to the upward force to find its value: \(F_{upward} = F_{downward} = 806 \mathrm{N}\) Thus, the upward force of the water acting on the mattress and man is 806 N.

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Most popular questions from this chapter

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 \(t=4.00 \mathrm{s} ?\)
You grab a book and give it a quick push across the top of a horizontal table. After a short push, the book slides across the table, and because of friction, comes to a stop. (a) Draw an FBD of the book while you are pushing it. (b) Draw an FBD of the book after you have stopped pushing it, while it is sliding across the table. (c) Draw an FBD of the book after it has stopped sliding. (d) In which of the preceding cases is the net force on the book not equal to zero? (e) If the book has a mass of \(0.50 \mathrm{kg}\) and the coefficient of friction between the book and the table is \(0.40,\) what is the net force acting on the book in part (b)? (f) If there were no friction between the table and the book, what would the free-body diagram for part (b) look like? Would the book slow down in this case? Why or why not?
A crate of oranges weighing \(180 \mathrm{N}\) rests on a flatbed truck $2.0 \mathrm{m}$ from the back of the truck. The coefficients of friction between the crate and the bed are \(\mu_{\mathrm{s}}=0.30\) and $\mu_{\mathrm{k}}=0.20 .\( The truck drives on a straight, level highway at a constant \)8.0 \mathrm{m} / \mathrm{s} .$ (a) What is the force of friction acting on the crate? (b) If the truck speeds up with an acceleration of \(1.0 \mathrm{m} / \mathrm{s}^{2},\) what is the force of the friction on the crate? (c) What is the maximum acceleration the truck can have without the crate starting to slide?
An engine pulls a train of 20 freight cars, each having a mass of $5.0 \times 10^{4} \mathrm{kg}$ with a constant force. The cars move from rest to a speed of \(4.0 \mathrm{m} / \mathrm{s}\) in \(20.0 \mathrm{s}\) on a straight track. Ignoring friction, what is the force with which the 10th car pulls the 11th one (at the middle of the train)? (school bus)
Luke stands on a scale in an elevator that has a constant acceleration upward. The scale reads \(0.960 \mathrm{kN} .\) When Luke picks up a box of mass $20.0 \mathrm{kg},\( the scale reads \)1.200 \mathrm{kN} .$ (The acceleration remains the same.) (a) Find the acceleration of the elevator. (b) Find Luke's weight.
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