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Vanyuwa [196]
3 years ago
5

Two forces act at a point in the plane. The angle between the two forces is given. Find the magnitude of the resultant force. fo

rces of 232 and 194 ​newtons, forming an angle of 67
Physics
1 answer:
Zielflug [23.3K]3 years ago
4 0

Answer:

408N at 89.89°

Explanation:

This problem requires that we resolve the force vectors into

x- and y

-componentsOnce this is done, we can add the components easily, as the one 2-dimensional problem will be two 1-dimensional problems.

Finally, we will convert the resultant force into standard form and find the equilibrant.

Resolve into components:

F1x =F1cos 180°= 232(−1)=−232N

F1y=F1sin180°=0N

F2x=F2cos(−140°)=194(−0.766)=−148.6N

F1y=F1sin(−140°)=232(−0.643)=−149.17N

Note the change of the angle used to give the direction of

F2. Standard angles (rotation from thex

-axis; counterclockwise is +) should be used to avoid sign errors in the results.

Now, we add the components:

Fx=F1x+F2x=−380.6N

Fy=F1y+F1y=−148.17N

Technically, this is the resultant force. However, it should be changed back into standard form. Here's how:

F=√(Fx)2(Fy)2=√(−380.6)^2(−148.17)^2=408N

θ=tan−1(−148.17−380.6)

=89.89°

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Explanation:

Let the left end be the reference end with string p closest to it.

Let CCW moments be positive

Sum moments about p to zero

1(9.8)[2 - 1] + Fq[6 - 2] - 5(9.8)[8/2 - 2] - 1.5(9.8)[5 - 2] - 2(9.8)[7 - 2] = 0

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Sum moments about q to zero

1(9.8)[6 - 1]  - Fp[6 - 2] + 5(9.8)[6 - 8/2] + 1.5(9.8)[6 - 5] - 2(9.8)[7 - 6] = 0

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Sum vertical forces to zero

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Fp = 35.525  ≈ 36 N

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