Forces of 11.5 N north, 20.4 N east, and 15.7 N south are simultaneously applied to a 3.74 kg mass as it rests on an air table. What is the magnitude of its acceleration?
Total Force = mass*acceleration
Add the forces together by adding the horizontal and vertical components individual and then using the Pythagorean theorem to find the hypotenuse (a.k.a total force)
For the vertical components, let North be +, while South be -. For the horizontal component, let East be +.
vertical component (y):
11.5N -15.7N
horizontal component (x):
20.4 N
total force:
sqrt[(11.5N -15.7N)^2 + (20.4N)^2]
now plug it into newton's second law equation (see above)
To find the magnitude of the acceleration, we need to calculate the net force acting on the object using the given forces.
First, let's break down the given forces into their respective components:
Force in the North direction (F₁) = 11.5 N
Force in the East direction (F₂) = 20.4 N
Force in the South direction (F₃) = 15.7 N
Since the forces are given in different directions, we need to resolve them into two perpendicular components, usually referred to as the x-component and y-component.
The x-component is in the East direction, so we can assign it a positive sign.
The y-component is in the North direction, so we can assign it a positive sign.
Now let's calculate the x and y components of the given forces:
Component of F₁ in the x-direction (F₁x) = 0 N (since it is in the North direction)
Component of F₂ in the x-direction (F₂x) = 20.4 N
Component of F₃ in the x-direction (F₃x) = 0 N (since it is in the South direction)
Component of F₁ in the y-direction (F₁y) = 11.5 N
Component of F₂ in the y-direction (F₂y) = 0 N (since it is in the East direction)
Component of F₃ in the y-direction (F₃y) = -15.7 N
Now, let's find the net force in the x and y directions separately:
Net force in the x-direction (F_net_x) = F₁x + F₂x + F₃x = 0 N + 20.4 N + 0 N = 20.4 N
Net force in the y-direction (F_net_y) = F₁y + F₂y + F₃y = 11.5 N + 0 N - 15.7 N = -4.2 N
Next, we can calculate the net force vector (F_net) using the components:
F_net = √(F_net_x² + F_net_y²) = √(20.4 N)² + (-4.2 N)² = √(416.16 N² + 17.64 N²) = √433.8 N² = 20.8 N
The magnitude of the acceleration (a) is equal to the net force divided by the mass of the object:
a = F_net / m = 20.8 N / 3.74 kg ≈ 5.56 m/s²
Therefore, the magnitude of the acceleration is approximately 5.56 m/s².
To find the magnitude of acceleration, we can use Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.
In this problem, we have three forces acting in different directions: 11.5 N north, 20.4 N east, and 15.7 N south. To find the net force, we need to find the resultant of these forces.
To find the resultant, we can break down the forces into their x and y components using trigonometry. The north and south forces are along the y-axis, while the east force is along the x-axis.
Given:
Force north (Fnorth) = 11.5 N
Force east (Feast) = 20.4 N
Force south (Fsouth) = 15.7 N
Mass (m) = 3.74 kg
To find the x-component of the resultant force (Rx), we subtract the east force from the west force:
Rx = Feast - Fsouth
= 20.4 N - 15.7 N
= 4.7 N
To find the y-component of the resultant force (Ry), we subtract the south force from the north force:
Ry = Fnorth - Fsouth
= 11.5 N - 15.7 N
= -4.2 N
Since the y-component is negative, it means that the net force is pointing in the downward direction, or south.
Now, we can calculate the net force (R) using the Pythagorean theorem:
R = √(Rx^2 + Ry^2)
= √(4.7^2 + (-4.2)^2)
= √(22.09 + 17.64)
= √39.73
≈ 6.30 N
Finally, we can find the magnitude of acceleration (a) using Newton's second law:
a = R / m
= 6.30 N / 3.74 kg
≈ 1.68 m/s²
Therefore, the magnitude of the mass's acceleration is approximately 1.68 m/s².