There are two forces on the 2.36 kg box in the overhead view of Fig. 5-31 but only one is shown. For F1 = 18.6 N, a = 11.4 m/s2, and θ = 26.7°, find the second force (a) in unit-vector notation and as (b) a magnitude and (c) a direction. (State the direction as a negative angle measured from the +x direction.)
To find the second force on the box, we can use vector addition. Let's denote the second force as F2.
(a) To find F2 in unit-vector notation, we can break down F1 and F2 into their x and y components.
The x-component of F1 is given by: F1x = F1 * cos(θ)
where θ is the angle between F1 and the x-axis.
Substituting the given values: F1x = 18.6 N * cos(26.7°)
The y-component of F1 is given by: F1y = F1 * sin(θ)
Substituting the given values: F1y = 18.6 N * sin(26.7°)
Now, let's express F2 in terms of its x and y components: F2 = F2x * i + F2y * j
Since the box is in equilibrium (net force is zero), the x-components and y-components of the forces must balance:
F1x + F2x = 0
F1y + F2y = 0
From these equations, we can solve for F2x and F2y:
F2x = -F1x
F2y = -F1y
(b) To find the magnitude of F2, we can use the Pythagorean theorem:
|F2| = sqrt(F2x^2 + F2y^2)
Substituting the values we found in part (a): |F2| = sqrt((-F1x)^2 + (-F1y)^2)
(c) To find the direction of F2, we can use trigonometry.
The direction angle of F2, measured in relation to the positive x-axis, can be found as:
θ = atan(F2y / F2x)
Substituting the values we found in part (a): θ = atan((-F1y) / (-F1x))
The direction angle will be negative since it is measured from the +x direction.
To summarize:
(a) F2 = -F1 * cos(θ) * i - F1 * sin(θ) * j
(b) |F2| = sqrt((-F1 * cos(θ))^2 + (-F1 * sin(θ))^2)
(c) θ = atan((-F1 * sin(θ)) / (-F1 * cos(θ)))
To find the second force on the box, we can use Newton's second law of motion. According to this law, the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
Let's break down the given information:
- F1 = 18.6 N is the magnitude of the first force (F1).
- a = 11.4 m/s^2 is the acceleration (a) of the box.
- θ = 26.7° is the angle between the first force (F1) and the x-axis.
First, let's find the x and y components of the first force (F1). We can use trigonometry to do this.
The x-component of F1 can be found using the formula:
F1x = F1 * cos(θ)
Substituting the given values:
F1x = 18.6 N * cos(26.7°)
Next, we can find the y-component of F1 using the formula:
F1y = F1 * sin(θ)
Substituting the given values:
F1y = 18.6 N * sin(26.7°)
Now, let's calculate the x and y components of the net force (Fnet).
The x-component of Fnet is the sum of the x-components of F1 and the second force:
Fnetx = F1x + F2x
The y-component of Fnet is the sum of the y-components of F1 and the second force:
Fnety = F1y + F2y
Since we are given the mass of the box (2.36 kg) and its acceleration (11.4 m/s^2), we can find Fnet using the formula:
Fnet = m * a
Now we can solve for the second force (F2).
Using the equation Fnetx = F1x + F2x, we can isolate F2x:
F2x = Fnetx - F1x
Using the equation Fnety = F1y + F2y, we can isolate F2y:
F2y = Fnety - F1y
To express the second force (F2) in unit-vector notation, we can write it as:
F2 = F2x * î + F2y * ĵ
To find the magnitude of the second force, we can use the Pythagorean theorem:
|F2| = sqrt(F2x^2 + F2y^2)
Finally, to find the direction of the second force, we can use the inverse tangent (arctan) function:
Direction = arctan(F2y / F2x)
The angle will be measured as a negative angle from the +x direction.
a)
Fnet = ma
F1 + F2 = ma
18.6N + F2 = 2.36(11.4*cos26.7° + 11.4*sin26.7°)
F2 = 42.6i + 12.1j
b)
sqrt(42.6^2 +12.1^2)
= 44.3 (N)
c)
θ = 26.7° + 180°
θ = 206.7°
206.7° - 300°
θ = -93.3°
a)
Fnet = ma
F1 + F2 = ma
18.6N + F2 = 2.36(11.4*cos26.7° + 11.4*sin26.7°)
F2 = 42.6i + 12.1j
b)
sqrt(42.6^2 +12.1^2)
= 44.3 (N)
c)
θ = 26.7° + 90°
θ = -116.7°