Two forces are applied to a car in an effort to move it, F1 is 400N and F2 is 450N, F1 is 30 degrees in the upper right and F2 is 10 degrees in the upper left)
(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.
All angles are = measured CCW from +x-axis.
a. Fr = 400 N.[30o] + 450N.[170] = Resultant force.
X = 400*Cos30+450*Cos170 = -97 N.
Y = 400*sin30+450*sin170 = 278 N.
Fr = -97 + 278i = 294N.[-71o] = 294N.[71o] S. of E. = 294N.[289o] CCW.
b. Fr = M*a = 294.
3000*a = 294,
a =
To find the resultant vector of two forces, we need to calculate the horizontal and vertical components of each force and then add them together to determine the resultant vector.
Let's break down the given information:
Force F1 = 400N, angle θ1 = 30 degrees (upper right)
Force F2 = 450N, angle θ2 = 10 degrees (upper left)
(a) To find the resultant vector, we need to calculate its horizontal and vertical components.
For Force F1:
Horizontal component of F1 = F1 * cos(θ1)
Vertical component of F1 = F1 * sin(θ1)
For Force F2:
Horizontal component of F2 = F2 * cos(θ2)
Vertical component of F2 = F2 * sin(θ2)
Let's calculate these components:
Horizontal component of F1 = 400N * cos(30°) = 400N * 0.866 = 346.4N (rounded to one decimal place)
Vertical component of F1 = 400N * sin(30°) = 400N * 0.5 = 200N
Horizontal component of F2 = 450N * cos(10°) = 450N * 0.9848 = 443.2N (rounded to one decimal place)
Vertical component of F2 = 450N * sin(10°) = 450N * 0.1736 = 78.12N (rounded to two decimal places)
Now, let's find the resultant vector by adding the horizontal and vertical components:
Horizontal component of resultant vector = Horizontal component of F1 + Horizontal component of F2
Vertical component of resultant vector = Vertical component of F1 + Vertical component of F2
Horizontal component of resultant vector = 346.4N + 443.2N = 789.6N
Vertical component of resultant vector = 200N + 78.12N = 278.12N
Therefore, the resultant vector of the two forces is approximately 789.6N to the right and 278.12N upwards.
(b) To calculate the acceleration of the car, we need to use Newton's second law, which states that force is equal to mass multiplied by acceleration (F = ma).
Given:
Mass of the car (m) = 3,000 kg
We know the magnitude of the resultant force from part (a), which is approximately equal to the magnitude of F1 + F2.
Magnitude of resultant force = √(horizontal component of resultant vector)^2 + (vertical component of resultant vector)^2
Magnitude of resultant force = √(789.6N)^2 + (278.12N)^2 ≈ √(622885.76N^2 + 77312.3744N^2) ≈ √(700198.1344N^2) ≈ 836.4N (rounded to one decimal place)
Now, we can use Newton's second law to calculate the acceleration:
F = ma
836.4N = 3,000kg * a
a = 836.4N / 3,000kg ≈ 0.2791 m/s^2 (rounded to four decimal places)
Therefore, the car has an acceleration of approximately 0.2791 m/s^2.