chapter 5 forces in two dimensions practice problems
lesson 1 vectors in two dimensions
example problem 1 - finding the magnitude of the sum of two vectors
1. you and your family are out for a drive. you drive 125.0 km due west, then turn due south and drive for another 65.0 km. what is the magnitude of your displacement? solve this problem both graphically and mathematically and check your answers against each other.
2. on a fine, sunny day, you and your siblings decide to go for a nearby hike. you walk 4.5 km in one direction, then make a 45° turn to the right and walk another 6.4 km. what is the magnitude of your displacement?
3. after picking up school supplies, you and your caregiver walk from the door of the mall to the car. you first walk 250.0 m down a lane of cars, and then turn 90° to the right and walk an additional 60.0 m. how far is the car from the door to the mall? solve this problem both graphically and mathematically and check your answers against each other.
4. sudhir walks 0.40 km in a direction 60.0° west of north then goes 0.50 km due west. what is his displacement?
5. you first walk 8.0 km north from home then walk east until your displacement from home is 10.0 km. how far east did you walk?
6. in a coordinate system in which the positive x - axis is east, for what range of angles is the x - component positive? for what range is it negative?
7. could a vector ever be shorter than one of its components? could a vector ever be equal in length to one of its components? explain.
8. two ropes of equal length are tied to a tree branch and hold up a childs swing. the tension in each rope is 2.28 n. what is the combined force (magnitude and direction) of the two ropes on the swing?

Question

Accepted Answer

1.
# Explanation:
## Step1: Identify the vectors as perpendicular
The two - displacement vectors (west and south) are perpendicular. We can use the Pythagorean theorem. Let the west - ward displacement $d_1 = 125.0$ km and the south - ward displacement $d_2=65.0$ km.
The magnitude of the displacement $d$ is given by $d=\sqrt{d_1^{2}+d_2^{2}}$.
## Step2: Calculate the magnitude
$d=\sqrt{(125.0)^{2}+(65.0)^{2}}=\sqrt{15625 + 4225}=\sqrt{19850}\approx140.9$ km
# Answer:
The magnitude of the displacement is approximately $140.9$ km.

2.
# Explanation:
## Step1: Use the law of cosines
Let $a = 4.5$ km, $b = 6.4$ km and $	heta=45^{\circ}$. The magnitude of the displacement $R$ is given by $R=\sqrt{a^{2}+b^{2}-2ab\cos(180 - 	heta)}=\sqrt{a^{2}+b^{2}+2ab\cos	heta}$.
## Step2: Substitute values and calculate
$R=\sqrt{(4.5)^{2}+(6.4)^{2}+2	imes4.5	imes6.4	imes\cos45^{\circ}}$
$=\sqrt{20.25 + 40.96+2	imes4.5	imes6.4	imes\frac{\sqrt{2}}{2}}$
$=\sqrt{20.25 + 40.96+(57.6	imes\frac{\sqrt{2}}{2})}$
$=\sqrt{20.25 + 40.96 + 40.73}\approx\sqrt{101.94}\approx10.1$ km
# Answer:
The magnitude of the displacement is approximately $10.1$ km.

3.
# Explanation:
## Step1: Recognize perpendicular vectors
The two - displacement vectors (along the lane of cars and then the perpendicular turn) are perpendicular. Let the first displacement $x = 250.0$ m and the second displacement $y = 60.0$ m.
Using the Pythagorean theorem, the distance $D$ from the mall door to the car is $D=\sqrt{x^{2}+y^{2}}$.
## Step2: Calculate the distance
$D=\sqrt{(250.0)^{2}+(60.0)^{2}}=\sqrt{62500+3600}=\sqrt{66100}\approx257.1$ m
# Answer:
The car is approximately $257.1$ m from the door of the mall.

4.
# Explanation:
## Step1: Resolve the first vector
The first vector of magnitude $A = 0.40$ km in the direction $60^{\circ}$ west of north. Its $x$ - component $A_x=- 0.40\sin60^{\circ}$ and $y$ - component $A_y = 0.40\cos60^{\circ}$. The second vector of magnitude $B = 0.50$ km has $x$ - component $B_x=-0.50$ km and $y$ - component $B_y = 0$.
The total $x$ - component $R_x=A_x + B_x=-0.40\sin60^{\circ}-0.50=-0.40	imes\frac{\sqrt{3}}{2}-0.50\approx - 0.346 - 0.50=-0.846$ km
The total $y$ - component $R_y=A_y + B_y=0.40\cos60^{\circ}+0 = 0.20$ km.
## Step2: Calculate the magnitude of the resultant
The magnitude of the displacement $R=\sqrt{R_x^{2}+R_y^{2}}=\sqrt{(-0.846)^{2}+(0.20)^{2}}=\sqrt{0.716 + 0.04}=\sqrt{0.756}\approx0.87$ km
# Answer:
The displacement is approximately $0.87$ km.

5.
# Explanation:
## Step1: Use the Pythagorean theorem
Let the north - ward displacement $y = 8.0$ km and the east - ward displacement $x$ and the resultant displacement $R = 10.0$ km.
By the Pythagorean theorem $R^{2}=x^{2}+y^{2}$, so $x=\sqrt{R^{2}-y^{2}}$.
## Step2: Calculate the east - ward displacement
$x=\sqrt{(10.0)^{2}-(8.0)^{2}}=\sqrt{100 - 64}=\sqrt{36}=6.0$ km
# Answer:
You walked $6.0$ km east.

6.
# Explanation:
## Step1: Recall the formula for the x - component of a vector
The $x$ - component of a vector $\vec{v}$ with magnitude $v$ and direction $	heta$ (measured counter - clockwise from the positive $x$ - axis) is $v_x=v\cos	heta$.
The $x$ - component is positive when $\cos	heta>0$. This occurs for $- 90^{\circ}<	heta<90^{\circ}$.
The $x$ - component is negative when $\cos	heta<0$. This occurs for $90^{\circ}<	heta<270^{\circ}$.
# Answer:
The $x$ - component is positive for $-90^{\circ}<	heta<90^{\circ}$ and negative for $90^{\circ}<	heta<270^{\circ}$.

7.
# Brief Explanations:
A vector cannot be shorter than one of its components because, by the Pythagorean theorem (for two - dimensional vectors) or the general formula for the magnitude of a vector in $n$ - dimensions, the magnitude of a vector $\vec{R}=\sqrt{R_1^{2}+R_2^{2}+\cdots+R_n^{2}}$. The magnitude is always greater than or equal to the absolute value of each individual component. A vector can be equal in length to one of its components when all other components are zero. For example, in two - dimensions, if $\vec{R}=(R_x,0)$ then $|\vec{R}| = |R_x|$.
# Answer:
A vector cannot be shorter than one of its components. A vector can be equal in length to one of its components when all other components are zero.

8.
# Explanation:
## Step1: Analyze the symmetry
Since the two ropes have equal tension $T = 2.28$ N and are symmetrically placed (assuming vertical equilibrium for the swing), the horizontal components of the tension in the two ropes cancel out. The vertical components add up.
The combined vertical force $F_y=2T$ (assuming the ropes are symmetrically placed vertically).
The magnitude of the combined force $F = 2T=2	imes2.28 = 4.56$ N and the direction is vertically upward.
# Answer:
The magnitude of the combined force is $4.56$ N and the direction is vertically upward.

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QUESTION IMAGE

Question

Explanation:

Step1: Identify the vectors as perpendicular

Step2: Calculate the magnitude

Step1: Use the law of cosines

Step2: Substitute values and calculate

Step1: Recognize perpendicular vectors

Step2: Calculate the distance

Answer: