Physics, 29.11.2019 10:31 darkdestroyer0888

Three forces of magnitudes f1=4.0n, f2=6.0n, and f3=8.0n are applied to a block of mass m=2.0kg, initially at rest, at angles shown on the diagram. (figure 1) in this problem, you will determine the resultant (net) force by combining the three individual force vectors. all angles should be measured counterclockwise from the positive x axis (i.
e., all angles are positive).

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*As seen in figure, we first find the resultant along x-axis and along y-axis and then find the net resultan force.

F(X) = F1(X) + F2(X) + F3(X)
= F1 cos25 + F2cos325 + F3 cos 180
= 4 cos 25 + 6 cos 325 + 8 cos 180
= .54 N

F(Y) = F1(Y) + F2(Y) + F3(Y)
= F1 sin25 + F2 sin325 + F3 sin 180
= 4 sin 25 + 6 sin 325 + 8 sin 180
= -1.75 N
Now,
F net = $\sqrt{ F(X)^{2} + F(Y)^{2} }$
= $\sqrt{ (0.54)^{2} + (-1.75)^{2} }$
= 1.8N
Direction Ф = $tan( \frac{1.75}{0.54}) ^{-1}$
= 73°
from x-axis - 360°-73°=287°
Three forces of magnitudes f1=4.0n, f2=6.0n, and f3=8.0n are applied to a block of mass m=2.0kg, ini

Three forces of magnitudes f1=4.0n, f2=6.0n, and f3=8.0n are applied to a block of mass m=2.0kg, ini

Answer from: jalenm266

The resultant force can be found by finding the sum of the vectors of the individual forces

The resultant (net) force vector and magnitude of the resultant of the three forces in the question are

The resultant force, $\overset \rightarrow F_r$ ≈ 0.54·i - 1.75·jThe magnitude of the resultant force is approximately 1.83142 N (Direction of approximately 287.15° CCW)

The reason the above values re correct is as follows:

The given forces, are;

F₁ = 4.0 N, direction; 25° North of East

F₂ = 6.0 N, direction; 325° counterclockwise measured from the x-axis = 35° South of East

F₃ = 8.0 N, direction; 180° counterclockwise = West

The representation of the forces in vector format is as follows;

$\overset \rightarrow F_1$ = 4 × cos(25°)·i + 4 × sin(25°)·j

$\overset \rightarrow F_2$ = 6 × cos(325°)·i - 6 × sin(325°)·j

$\overset \rightarrow F_3$ = -8 × cos(180°)·i + 8 × sin(180°)·j

The resultant force, $\overset \rightarrow F_r$ is given as follows;

$\overset \rightarrow F_r$ = $\mathbf{\overset \rightarrow F_1}$ + $\mathbf{\overset \rightarrow F_2}$ + $\mathbf{\overset \rightarrow F_3}$

∴ $\overset \rightarrow F_r$ = (4 × cos(25°) + 6 × cos(325°) - 8 × cos(180°))·i + ((6 × sin(25°) - 6 × sin(325°) + 8 × sin(180°))·j

Which gives;

$\overset \rightarrow F_r$ = (4 × cos(25°) + 6 × cos(325°) - 8)·i + ((6 × sin(25°) + 6 × sin(325°))·j

$\overset \rightarrow F_r$ = 0.54·i - 1.75·j

The resultant force, $\overset \rightarrow F_r$ = 0.54·i - 1.75·j

The direction of the resultant force, θ = arctan((- 1.75)/0.54) ≈ - 72.85° ≈ 287.15° Counterclockwise from the positive x-axis direction

The magnitude of the resultant force, $| \overset \rightarrow F_r |$ = √(0.54² + 1.75²) ≈ 1.83142

The magnitude of the resultant force, $| \overset \rightarrow F_r |$ ≈ 1.83142 N

Learn more about force vectors here:

link

Answer from: Quest

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Answer from: Quest

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Three forces of magnitudes f1=4.0n, f2=6.0n, and f3=8.0n are applied to a block of mass m=2.0kg, ini...

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