I need the answers
Two forces are applied to a car in an effort to move it, as shown in the following figure, where F1 = 453 N and F2 = 400 N. (Assume up and to the right as positive directions.)(F1 is 8degrees and F2 is 35degreed)
(a) What is the resultant of these two forces?
magnitude
direction
(b) If the car has a mass of 3805 kg, what acceleration does it have? Ignore friction.
a. Fr = 453N.[8o] + 400N.[35o].
Fr = x+yi = (453*Cos8+ 400*Cos35) + (453*sin8+400*sin35)I.
Fr = 776.3 + 292.5i = 829.6N.[20.6o].
b. Fr = M*a.
a = Fr/M = 829.6[20.6]/3805 = 0.218m/s^2[20.6o].
To find the resultant of the two forces, we will need to use vector addition. Let's break down the given forces into their horizontal and vertical components:
Force F1:
Magnitude: 453 N
Direction: 8 degrees above the positive x-axis
Force F2:
Magnitude: 400 N
Direction: 35 degrees above the positive x-axis
(a) Resultant Force:
To find the resultant force, we need to add the horizontal and vertical components of the two forces separately:
Horizontal Component:
F1x = F1 * cos θ1
= 453 N * cos 8°
≈ 449.884 N
F2x = F2 * cos θ2
= 400 N * cos 35°
≈ 327.324 N
Total Horizontal Component:
Fx = F1x + F2x
≈ 449.884 N + 327.324 N
≈ 777.208 N
Vertical Component:
F1y = F1 * sin θ1
= 453 N * sin 8°
≈ 63.03 N
F2y = F2 * sin θ2
= 400 N * sin 35°
≈ 229.302 N
Total Vertical Component:
Fy = F1y + F2y
≈ 63.03 N + 229.302 N
≈ 292.332 N
Now, we can determine the magnitude and direction of the resultant force using these components:
Magnitude of Resultant Force:
Fr = √(Fx² + Fy²)
= √((777.208 N)² + (292.332 N)²)
≈ 836.94 N
Direction of Resultant Force:
θr = tan⁻¹(Fy / Fx)
= tan⁻¹(292.332 N / 777.208 N)
≈ 20.81° above the positive x-axis
Therefore, the resultant of these two forces has a magnitude of approximately 836.94 N and a direction approximately 20.81° above the positive x-axis.
(b) Acceleration of the car:
To find the acceleration of the car, we will use Newton's second law:
F = m * a
where F is the resultant force and m is the mass of the car.
Plugging in the given values:
F = 836.94 N
m = 3805 kg
a = F / m
= 836.94 N / 3805 kg
≈ 0.220 m/s²
Therefore, the car has an acceleration of approximately 0.220 m/s².
To find the resultant force of two forces, F1 and F2, you can use vector addition. The resultant force magnitude can be found using the Pythagorean Theorem, and the direction can be found using trigonometry.
(a) To find the magnitude of the resultant force, you can use the Pythagorean Theorem, which states that the square of the hypotenuse (R) of a right triangle is equal to the sum of the squares of the other two sides. In this case, F1 and F2 are the two sides, and R is the hypotenuse.
Let's calculate the magnitude of the resultant force:
R^2 = F1^2 + F2^2
R^2 = (453 N)^2 + (400 N)^2
R^2 = 205,209 N^2 + 160,000 N^2
R^2 = 365,209 N^2
Taking the square root of both sides, we find the magnitude of the resultant force of these two forces:
R = sqrt(365,209 N^2)
R ≈ 604.89 N
Therefore, the magnitude of the resultant force is approximately 604.89 N.
To find the direction of the resultant force, you can use trigonometry. We are given the angles of F1 and F2 with respect to the positive x-axis. Since F1 is at an angle of 8 degrees and F2 is at an angle of 35 degrees, we can use these angles to find the direction of the resultant force.
Let's calculate the direction of the resultant force:
tan θ = (F2y + F1y)/(F1x + F2x)
Where F1x and F1y are the x and y components of F1, and F2x and F2y are the x and y components of F2.
F1x = F1 * cos(8 degrees)
F1y = F1 * sin(8 degrees)
F2x = F2 * cos(35 degrees)
F2y = F2 * sin(35 degrees)
Plugging in the given values:
F1x = 453 N * cos(8 degrees)
F1y = 453 N * sin(8 degrees)
F2x = 400 N * cos(35 degrees)
F2y = 400 N * sin(35 degrees)
tan θ = (453 N * sin(8 degrees) + 400 N * sin(35 degrees))/(453 N * cos(8 degrees) + 400 N * cos(35 degrees))
Using a calculator, we can evaluate this expression:
tan θ ≈ 0.478
To find the direction, we can take the inverse tangent (arctan or tan^(-1)) of the result:
θ ≈ arctan(0.478)
θ ≈ 25.73 degrees
Therefore, the direction of the resultant force is approximately 25.73 degrees.
(b) To find the acceleration of the car, we can use Newton's second law of motion, which states that the force acting on an object is equal to the object's mass multiplied by its acceleration.
F = ma
In this case, the resultant force (R) is acting on the car, and its mass (m) is given as 3805 kg. We can rearrange the equation to solve for the acceleration (a):
a = F/m
Plugging in the values we previously calculated:
a = (604.89 N)/(3805 kg)
a ≈ 0.159 m/s^2
Therefore, the car has an acceleration of approximately 0.159 m/s^2.