Question

the force resultant P is less than the resultant Q then the resultant is
a) along the bisector
b)closer to p
c)closer to p
d) out side of two forces ​

Answer 1

The force resultant p is a smaller amount than the resultant letter of the alphabet then the resultant is closer to p.

Parallelogram Law of Forces:

• If 2 forces, functioning at a degree be described in magnitude and direction by the 2 adjacent sides of a quadrilateral, then their resultant is described in magnitude and direction by the diagonal of the quadrilateral passing through that time.

Application for Actual Alignment

• The unit resultant force curve is calculated beneath the condition of the tangent level track, namely, icon = 0.

• If there are rea unit grades, as well as equivalent curve grades or equivalent tunnel grades, it's necessary to subtract the regenerate extra resistances from the unit resultant force, namely c = f(V) − 10icon.

• therefore simply move the vertical axis of the resultant force diagram by an icon value.

• If this is positive, the vertical axis is stirred to the left; if the other n is negative, the vertical axis is stirred to the correct.

• Then the c = f(V) curves within the new axis re-refer the unit resultant force curve once trade-in runs on the icon grade.

The resultant of 2 forces P and let the ter of the alphabet is R

R = P + Q

|R| = sqrt(P^2 + Q^2 + 2PQCosθ)

When leta ter of the alphabet is doubled new resultant R’ is

R’ = P + 2Q

Since it's perpendicular tin then their product of R’ and P is zero

(R’). (P) = 0

(P + 2Q). (P) = 0

P^2 + 2PQCosθ = zero

(Dot product of P and P is P^2 which of P and 2Q is 2PQCosθ)

Cosθ = -P/(2Q)

R = sqrt(P^2 + Q^2 + 2PQCosθ)

R = sqrt(P^2 + Q^2 + 2PQ*(-P/(2Q))

R = sqrt(P^2 + Q^2 - P^2)

R = Q

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Answer 2

1. If the magnitude and direction of two forces acting at a degree are described by the two adjacent sides of a quadrilateral, then the resultant is described by the diagonal of the quadrilateral.
2. So when the force resultant $P$ is less than the resultant $Q$ , then the resultant is closer to $P$.

Hence, option B is correct.

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