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An illustration shows an overhead view of a car as it approaches a fork in the road. The car is oriented such that it is directed toward the top of the illustration. A force F1 extends from the nose of the car and points along the left-side forked road at an angle of 10° measured counterclockwise from the car's initial direction. A force F2 extends from the nose of the car and points along the right-side forked road at an angle of 30° measured clockwise from the car's initial direction. Two forces are applied to a car in an effort to move it, as shown in the figure below. (Let F1 = 440 N and F2 = 366 N. Assume up and to the right are in the positive directions.) (a) What is the resultant vector of these two forces? (b) If the car has a mass of 3,000 kg, what acceleration does it have? Ignore friction.

Question
An illustration shows an overhead view of a car as it approaches a fork in the road. The car is oriented such that it is directed toward the top of the illustration. A force F1 extends from the nose of the car and points along the left-side forked road at an angle of 10° measured counterclockwise from the car's initial direction. A force F2 extends from the nose of the car and points along the right-side forked road at an angle of 30° measured clockwise from the car's initial direction.
 
Two forces are applied to a car in an effort to move it, as shown in the figure below. (Let 
F1 = 440 N
 and 
F2 = 366 N.
 Assume up and to the right are in the positive directions.)
 
(a) What is the resultant vector of these two forces?
 
(b) If the car has a mass of 3,000 kg, what acceleration does it have? Ignore friction.
check_circleAnswer
Step 1

Draw the components of the forces F1  and F2.

=440N
FLOS 30
Fieio
366N
10
30
Fi sin to
F2sin 30
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=440N FLOS 30 Fieio 366N 10 30 Fi sin to F2sin 30

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Step 2

Write the x and y components of the force F1  and F2 from the diagram.

components of F along x and y direction from the diagram,
FR =-F sin10°
= -440 N (sin10°)
=-76.40 N
FFcos10
1у
- 440 Ncos10°
-433.315 N
components of F, along x and y direction from the diagram
Fir F sin 30°
- 366 N (sin30°)
2x
= 183 N
Fy F,cos30
= 366 Ncos30°
= 316.96N
help_outline

Image Transcriptionclose

components of F along x and y direction from the diagram, FR =-F sin10° = -440 N (sin10°) =-76.40 N FFcos10 1у - 440 Ncos10° -433.315 N components of F, along x and y direction from the diagram Fir F sin 30° - 366 N (sin30°) 2x = 183 N Fy F,cos30 = 366 Ncos30° = 316.96N

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Step 3

Continuing previ...

Sum of forces on the x direction
FFF
=(-76.40+183)N
106.6N
Sum of forces on the y direction
F FFy
+
-(433.315+316.96)N
= 750.27 N
Resultant vector of two forces is F (106.6N)i +(750.27 N)
E.-Σ.)- Ση
Net
(106.6N) +(750.27N)
757.80 N
Magnitude of resulatant vector is Fet 757.80 N
Σ .
Ση
tan
Σ .
ΣΗ
0=tan
750.27N
- tan
106.6N
81.91°
WW
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Sum of forces on the x direction FFF =(-76.40+183)N 106.6N Sum of forces on the y direction F FFy + -(433.315+316.96)N = 750.27 N Resultant vector of two forces is F (106.6N)i +(750.27 N) E.-Σ.)- Ση Net (106.6N) +(750.27N) 757.80 N Magnitude of resulatant vector is Fet 757.80 N Σ . Ση tan Σ . ΣΗ 0=tan 750.27N - tan 106.6N 81.91° WW

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Physics

Newtons Laws of Motion

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