Four forces are acting on an object in static equilibrium: Force 1 has a magnitude of 46.0 N and is acting at an angle of (54) degrees above the positive x-axis. Force 2 has a magnitude of (42) N and is acting along the positive x-axis. Force 3 has a magnitude of 34.5 N and is acting along the negative y-axis. Find the magnitude of the last force, F4. Give your answer in newtons (N) and with 3 significant figures.
To find the magnitude of the last force (F4), we can use the concept of vector addition. In static equilibrium, the sum of all forces acting on an object is equal to zero.
Let's break down the given information:
Force 1 has a magnitude of 46.0 N and is acting at an angle of 54 degrees above the positive x-axis. We can represent this force as follows:
F1 = 46.0 N at an angle of 54 degrees.
Force 2 has a magnitude of 42.0 N and is acting along the positive x-axis:
F2 = 42.0 N along the positive x-axis.
Force 3 has a magnitude of 34.5 N and is acting along the negative y-axis:
F3 = 34.5 N along the negative y-axis.
Now, we need to find the components of Force 1 in the x and y directions. To do this, we can use trigonometry.
The x-component of Force 1 can be found using the cosine function:
Fx1 = F1 * cos(angle)
= 46.0 N * cos(54 degrees)
Similarly, the y-component of Force 1 can be found using the sine function:
Fy1 = F1 * sin(angle)
= 46.0 N * sin(54 degrees)
Next, let's calculate the x and y components of all the forces:
Fx1 = 46.0 N * cos(54 degrees)
= 46.0 N * 0.5878
≈ 27.0 N (rounded to 3 significant figures)
Fy1 = 46.0 N * sin(54 degrees)
= 46.0 N * 0.8090
≈ 37.2 N (rounded to 3 significant figures)
Fx2 = 42.0 N (since it is along the positive x-axis, the y-component is zero)
Fy2 = 0 N (since it is along the positive x-axis)
Fx3 = 0 N (since it is along the negative y-axis)
Fy3 = -34.5 N (negative because it is along the negative y-axis)
Now, to find the magnitude of the last force (F4), we can sum up all the x and y components and set them equal to zero:
∑Fx = Fx1 + Fx2 + Fx3 + Fx4 = 0
∑Fy = Fy1 + Fy2 + Fy3 + Fy4 = 0
Since we are trying to find the magnitude of F4, we can represent it in terms of its components:
F4 = √(Fx4^2 + Fy4^2)
Using this information, we can solve for F4:
Fx4 = -Fx1 - Fx2 - Fx3
Fy4 = -Fy1 - Fy2 - Fy3
Plugging in the values we calculated earlier, we have:
Fx4 = -27.0 N
Fy4 = -37.2 N
Now, we can calculate the magnitude of F4:
F4 = √(Fx4^2 + Fy4^2)
= √((-27.0 N)^2 + (-37.2 N)^2)
≈ √(729 N^2 + 1380.84 N^2)
≈ √(2109.84 N^2)
≈ 45.931 N (rounded to 3 significant figures)
Therefore, the magnitude of the last force (F4) is approximately 45.931 N.
46N.[52o] + 29[0o] + 34.5[270o]+F4 = 0.
46*Cos52 + i46*sin52 + 29 -34.5i +F4=0
28.32 + 36.25i + 29 - 34.5i + F4 = 0.
57.32 + 1.75i + F4 = 0.
F4 = -57.32 - 1.75i.
X = -57.32 N., Y = -1.75 N.
F4 = sqrt(X^2 + Y^2).
Hope this helps.