Five forces act on an object.
(1) 56 N at 90° <-
(2) 40 N at 0°
(3) 75 N at 270° <-
(4) 40 N at 180°
(5) 50 N at 60°
What are the magnitude and direction of a sixth force that would produce equilibrium?
Answer to Find:
? N at ?°
To find the magnitude and direction of the sixth force that would produce equilibrium, we need to calculate the resultant force by adding all the given forces together and find the opposite force to balance the system.
Step 1: Resolve the forces into horizontal and vertical components.
Given forces:
(1) 56 N at 90° -> (horizontal component: 0, vertical component: 56 N)
(2) 40 N at 0° -> (horizontal component: 40 N, vertical component: 0)
(3) 75 N at 270° -> (horizontal component: 0, vertical component: -75 N)
(4) 40 N at 180° -> (horizontal component: -40 N, vertical component: 0)
(5) 50 N at 60° -> (horizontal component: 25 N, vertical component: 43.3 N)
Step 2: Sum up the horizontal and vertical components separately.
Horizontal component: 0 + 40 N + 0 - 40 N + 25 N = 25 N
Vertical component: 56 N + 0 - 75 N + 0 + 43.3 N = 24.3 N
Step 3: Calculate the magnitude of the resultant force.
Magnitude of the resultant force = √(horizontal component)^2 + (vertical component)^2
Magnitude = √(25 N)^2 + (24.3 N)^2
Magnitude ≈ 34.4 N
Step 4: Calculate the angle of the resultant force.
Angle = arctan(vertical component / horizontal component)
Angle = arctan(24.3 N / 25 N)
Angle ≈ 45.84°
Therefore, the magnitude of the sixth force that would produce equilibrium is approximately 34.4 N at an angle of approximately 45.84°.
To find the magnitude and direction of the sixth force that would produce equilibrium, we need to analyze the given forces and determine their resultant.
Step 1: Resolve forces into their horizontal and vertical components:
For forces (1), (3), and (5), we need to find their horizontal and vertical components.
Force (1):
Magnitude = 56 N
Direction = 90°
The horizontal component is 0 N (since it acts vertically) and the vertical component is 56 N.
Force (3):
Magnitude = 75 N
Direction = 270°
The horizontal component is 0 N (since it acts vertically) and the vertical component is -75 N (negative because it points downward).
Force (5):
Magnitude = 50 N
Direction = 60°
The horizontal component is 25 N (50 N * cos(60°)) and the vertical component is 43.3 N (50 N * sin(60°)).
Forces (2) and (4) are already in their horizontal components.
Force (2):
Magnitude = 40 N
Direction = 0°
The horizontal component is 40 N and the vertical component is 0 N (since it acts horizontally).
Force (4):
Magnitude = 40 N
Direction = 180°
The horizontal component is -40 N (negative because it acts in the opposite direction) and the vertical component is 0 N (since it acts horizontally).
Step 2: Calculate the magnitudes and directions of horizontal and vertical components:
Now, add the horizontal and vertical components separately:
Horizontal component = 40 N + (-40 N) + 0 N + 25 N + 0 N = 25 N
Vertical component = 56 N + (-75 N) + 43.3 N + 0 N + 0 N = 24.3 N
Step 3: Calculate the magnitude and direction of the resultant force:
To find the magnitude and direction of the resultant force, we can use the Pythagorean theorem and inverse tangent function.
Magnitude of the resultant force = sqrt((horizontal component)^2 + (vertical component)^2) = sqrt((25 N)^2 + (24.3 N)^2) ≈ 34.7 N
Direction of the resultant force = atan(vertical component / horizontal component) = atan(24.3 N / 25 N) = atan(0.972) ≈ 46° (rounded to the nearest whole degree)
Therefore, the magnitude of the sixth force that would produce equilibrium is approximately 34.7 N, and its direction is approximately 46°.
(2) and (4) cancel each other
(1) and (3) result in 19 N @ 270º
find the net resultant
find the cancelling (balancing) force
... equal magnitude , opposite direction
Fr = 56i + 40 - 75i -40 + (50*Cos60 + 50*sin60).
X = 40 - 40 + 50*Cos60 = 25N.
Y = 56 - 75 + 50*sin60 = -18.13 N.
Fr = sqrt(X^2 + Y^2) = 30.9 N.
TanA = Y/X, A = -35.95o = 35.95o S. of E. = 324o CCW.
The equilibrant force(Fe) must be equal in magnitude and 180o out phase with Fr.
Fe = 30.9N[35.95o] N. of W. = 30.9 N.[144.1o ] CCW.