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Chapter 4: Problem 6

Juan is helping his mother rearrange the living room furniture. Juan pushes on the armchair with a force of \(30 \mathrm{N}\) directed at an angle of \(15^{\circ}\) above a horizontal line while his mother pushes with a force of \(40 \mathrm{N}\) directed at an angle of \(20^{\circ}\) below the same horizontal. What is the vector sum of these two forces?

Short Answer

Expert verified
Answer: The vector sum of the two forces is approximately 66.743 N at an angle of 5.03° below the horizontal.

Step by step solution

01

Break each force into its horizontal and vertical components

First, we will find the horizontal and vertical components of each force. Let's start with Juan's force of 30 N at an angle of 15° above the horizontal: $$ F_{1x} = F_1 \cos(15^{\circ}) = 30 \cos(15^{\circ}) \approx 28.976 \mathrm{N} $$ $$ F_{1y} = F_1 \sin(15^{\circ}) = 30 \sin(15^{\circ}) \approx 7.794 \mathrm{N} $$ Now, we will find the components of his mother's force of 40 N at an angle of 20° below the horizontal: $$ F_{2x} = F_2 \cos(20^{\circ}) = 40 \cos(20^{\circ}) \approx 37.588 \mathrm{N} $$ $$ F_{2y} = -F_2 \sin(20^{\circ}) = -40 \sin(20^{\circ}) \approx -13.635 \mathrm{N} $$ Notice that the vertical component of the mother's force is negative since it is directed downwards.
02

Add the horizontal and vertical components

Now that we have the horizontal and vertical components of each force, we can add them together to find the total horizontal and vertical components of the resultant force: $$ F_{Rx} = F_{1x} + F_{2x} = 28.976 + 37.588 \approx 66.564 \mathrm{N} $$ $$ F_{Ry} = F_{1y} + F_{2y} = 7.794 - 13.635 \approx -5.841 \mathrm{N} $$
03

Find the magnitude and angle of the resultant force

Using the Pythagorean theorem, we can find the magnitude of the resultant force: $$ F_R = \sqrt{F_{Rx}^2 + F_{Ry}^2} = \sqrt{(66.564)^2 + (-5.841)^2} \approx 66.743 \mathrm{N} $$ To find the angle of the resultant force, we can use the arctangent function: $$ \theta = \arctan\left(\frac{F_{Ry}}{F_{Rx}}\right) = \arctan\left(\frac{-5.841}{66.564}\right) \approx -5.03^{\circ} $$ The negative angle means that the direction of the resultant force is 5.03° below the horizontal. Thus, the vector sum of the two forces is approximately 66.743 N at an angle of 5.03° below the horizontal.

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Most popular questions from this chapter

A barge is hauled along a straight-line section of canal by two horses harnessed to tow ropes and walking along the tow paths on either side of the canal. Each horse pulls with a force of \(560 \mathrm{N}\) at an angle of \(15^{\circ}\) with the centerline of the canal. Find the sum of the two forces exerted by the horses on the barge.
Margie, who weighs \(543 \mathrm{N},\) is standing on a bathroom scale that weighs \(45 \mathrm{N}\). (a) With what force does the scale push up on Margie? (b) What is the interaction partner of that force? (c) With what force does the Earth push up on the scale? (d) Identify the interaction partner of that force.
A hanging potted plant is suspended by a cord from a hook in the ceiling. Draw an FBD for each of these: (a) the system consisting of plant, soil, and pot; (b) the cord; (c) the hook; (d) the system consisting of plant, soil, pot, cord, and hook. Label each force arrow using subscripts (for example, \(\overrightarrow{\mathbf{F}}_{\mathrm{ch}}\) would represent the force exerted on the cord by the hook).
A bike is hanging from a hook in a garage. Consider the following forces: (a) the force of the Earth pulling down on the bike, (b) the force of the bike pulling up on the Earth, (c) the force of the hook pulling up on the bike, and (d) the force of the hook pulling down on the ceiling. Which two forces are equal and opposite because of Newton's third law? Which two forces are equal and opposite because of Newton's first law?
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