A particle P lies on a smooth horizontal surface. It is acted on by two horizontal forces of magnitudes 25 N at an angle of 30 degree with the horizontal and 20 N at an angle of 45 degree with the horizontal. Both forces acted away from the particle. A third force F is required to keep the particle in equilibrium.
Obtain F as a vector, giving its components to one decimal
To find the magnitude and direction of the third force F that is required to keep the particle in equilibrium, we can use the concept of vector addition.
Step 1: Draw a diagram.
Draw a horizontal line to represent the smooth surface. Place a dot on it to represent the particle P. Draw two arrows representing the two given forces: 25 N at an angle of 30 degrees and 20 N at an angle of 45 degrees. Note that both forces are acting away from the particle.
Step 2: Resolve each force into its x and y components.
Since the surface is horizontal, the y-components of both forces will cancel each other out, as there is no vertical motion. We only need to consider the x-components of the forces.
The x-component of the 25 N force at 30 degrees can be found using the formula:
Force_x = Force * cos(angle) = 25 N * cos(30 degrees) = 25 N * √3/2
Similarly, the x-component of the 20 N force at 45 degrees can be found using the formula:
Force_x = Force * cos(angle) = 20 N * cos(45 degrees) = 20 N * √2/2
Step 3: Add the x-components of the forces.
To find the x-component of the third force F, we need to add the x-components of the given forces:
F_x = (25 N * √3/2) + (20 N * √2/2)
Step 4: Find the y-component of the force F.
Since the particle is in equilibrium, the vertical forces should also balance each other out. Therefore, the y-component of the third force F should be zero.
Step 5: Write the vector components of the force F.
The vector components of the force F are:
F = (F_x, 0)
Step 6: Calculate the magnitude and direction of the force F.
The magnitude of the force F can be found using the Pythagorean theorem:
|F| = √(F_x^2 + F_y^2) = √(F_x^2 + 0^2) = √(F_x^2)
To find the direction of the force F, we can use the tangent function:
angle = arctan(F_y/F_x) = arctan(0/F_x) = arctan(0) = 0 degrees
Finally, substitute the value of F_x calculated in Step 3 to find the magnitude and direction of the force F.
To find the third force F required to keep the particle in equilibrium, we need to sum up all the horizontal forces acting on the particle.
Let's break down the given forces into their horizontal and vertical components:
Force 1 (25 N at an angle of 30 degrees with the horizontal):
Horizontal component = 25 N * cos(30°)
Vertical component = 25 N * sin(30°)
Force 2 (20 N at an angle of 45 degrees with the horizontal):
Horizontal component = 20 N * cos(45°)
Vertical component = 20 N * sin(45°)
Since the particle is in equilibrium, the sum of the horizontal forces acting on it must be zero. Therefore, the horizontal components of the forces should cancel each other out:
Force 1 (horizontal) + Force 2 (horizontal) + Force 3 (horizontal) = 0
(25 N * cos(30°)) + (20 N * cos(45°)) + (F (horizontal)) = 0
Now, let's solve for the horizontal component of the third force:
F (horizontal) = - [(25 N * cos(30°)) + (20 N * cos(45°))]
Using a calculator, we can compute the value of F (horizontal):
F (horizontal) ≈ - [21.65 + 14.14]
F (horizontal) ≈ - 35.8 N (rounded to one decimal place)
Since the third force F is acting away from the particle, its horizontal component will be negative.
Therefore, the horizontal component of the third force required to keep the particle in equilibrium is approximately -35.8 N.
As for the vertical component, since the particle is on a smooth horizontal surface, there is no vertical acceleration. Thus, the sum of the vertical forces acting on it must be zero:
Force 1 (vertical) + Force 2 (vertical) + Force 3 (vertical) = 0
(25 N * sin(30°)) + (20 N * sin(45°)) + (F (vertical)) = 0
Simplifying further, we have:
F (vertical) = - [(25 N * sin(30°)) + (20 N * sin(45°))]
Using a calculator, we can compute the value of F (vertical):
F (vertical) ≈ - [(25 × 0.5) + (20 × 0.7071)]
F (vertical) ≈ - [12.5 + 14.142]
F (vertical) ≈ - 26.642 N (rounded to one decimal place)
Since the third force F is acting away from the particle, its vertical component will be negative.
Therefore, the vertical component of the third force required to keep the particle in equilibrium is approximately -26.6 N.
Thus, the vector representation of the third force F is approximately F = (-35.8 N, -26.6 N)