Forces with magnitudes of 2000 newtons and 900 newtons act on the machine part at angles of 30 degrees and -45 degrees respectively, with the x-axis. Find the direction and magnitude of the resultant of these forces.
I see two vectors forming an angle of 75°
Draw the usual parallelogram with the resultant as the diagonal of that parallelogram.
R^2 = 900^2 + 2000^2 - 2(900)(2000)cos 105°
R = 2396
that's your magnitude
Now use the sine law to find the direction angle.
To find the direction and magnitude of the resultant of these forces, we can use vector addition.
Step 1: Resolve the forces into their x and y components.
The force with a magnitude of 2000 N at an angle of 30 degrees with the x-axis can be resolved into its x and y components as follows:
Fx1 = 2000 N * cos(30°)
Fy1 = 2000 N * sin(30°)
The force with a magnitude of 900 N at an angle of -45 degrees with the x-axis can be resolved into its x and y components as follows:
Fx2 = 900 N * cos(-45°)
Fy2 = 900 N * sin(-45°)
Step 2: Calculate the x and y components of the resultant force.
Rx = Fx1 + Fx2
Ry = Fy1 + Fy2
Step 3: Calculate the magnitude and direction of the resultant force.
Magnitude of the resultant force (R) can be calculated using the Pythagorean theorem:
R = sqrt(Rx^2 + Ry^2)
Direction of the resultant force (θ) can be calculated using the inverse tangent function:
θ = arctan(Ry / Rx)
Now let's calculate the values:
Fx1 = 2000 N * cos(30°) = 2000 N * (√3/2) = 2000√3/2 ≈ 1732 N
Fy1 = 2000 N * sin(30°) = 2000 N * (1/2) = 2000/2 = 1000 N
Fx2 = 900 N * cos(-45°) = 900 N * (√2/2) = 900√2/2 ≈ 318 N
Fy2 = 900 N * sin(-45°) = 900 N * (-√2/2) = -900√2/2 ≈ -318 N
Rx = Fx1 + Fx2 = 1732 N + 318 N = 2050 N
Ry = Fy1 + Fy2 = 1000 N + (-318 N) = 682 N
R = sqrt(Rx^2 + Ry^2) = sqrt((2050 N)^2 + (682 N)^2) = sqrt(4202500 N^2 + 465124 N^2) = sqrt(4667624 N^2) ≈ 2159 N
θ = arctan(Ry / Rx) = arctan(682 N / 2050 N) ≈ 18.77°
Therefore, the direction of the resultant force is approximately 18.77 degrees with the x-axis, and the magnitude of the resultant force is approximately 2159 N.
To find the direction and magnitude of the resultant of these forces, we can use vector addition.
First, we need to find the x and y components of each force. The x-component of a force is given by the formula:
Fx = F * cos(theta)
where F is the magnitude of the force and theta is the angle it makes with the x-axis.
For the 2000 N force:
Fx1 = 2000 N * cos(30°) = 2000 N * 0.866 = 1732 N (rounded to three decimal places)
For the 900 N force:
Fx2 = 900 N * cos(-45°) = 900 N * 0.707 = 636.364 N (rounded to three decimal places)
The y-component of a force is given by the formula:
Fy = F * sin(theta)
where F is the magnitude of the force and theta is the angle it makes with the x-axis.
For the 2000 N force:
Fy1 = 2000 N * sin(30°) = 2000 N * 0.5 = 1000 N
For the 900 N force:
Fy2 = 900 N * sin(-45°) = 900 N * -0.707 = -636.364 N (rounded to three decimal places)
Now, we can add the x and y components of the forces separately to find the resultant:
Rx = Fx1 + Fx2
= 1732 N + 636.364 N
= 2368.364 N (rounded to three decimal places)
Ry = Fy1 + Fy2
= 1000 N - 636.364 N
= 363.636 N (rounded to three decimal places)
To find the magnitude of the resultant (R), we can use the Pythagorean theorem:
R = sqrt(Rx^2 + Ry^2)
= sqrt((2368.364 N)^2 + (363.636 N)^2)
= sqrt(5600069.697 + 132144.697)
= sqrt(5732214.394)
= 2396.453 N (rounded to three decimal places)
Finally, let's find the direction of the resultant by calculating the angle it makes with the x-axis:
theta = arctan(Ry / Rx)
= arctan(363.636 N / 2368.364 N)
= arctan(0.1535)
= 8.768° (rounded to three decimal places)
Therefore, the direction of the resultant is approximately 8.768 degrees with the x-axis, and the magnitude of the resultant is approximately 2396.453 newtons.