Two forces P and Q have a resultant perpendicular to P The angle between the forces is

Question

Two forces P and Q have a resultant perpendicular to P The angle between the forces is (A)  tan -1((-P/Q)) (B)  tan -1((P/Q)) (C)  sin -1((P/Q)) (D)  cos -1(-(P/Q))

Tardigrade Admin · Accepted Answer

The angle made by resultant (R) of the forces P and Q with P is given by tan θ=(Q sin α/P+Q cos α) When θ=90°, then tan 90°=∞ ∴ P+Q cos α=0 or cos α=-(P/Q) or α= cos -1(-(P/Q))

Q. Two forces $P$ and $Q$ have a resultant perpendicular to $P$ The angle between the forces is

$tan^{- 1} (\frac{- P}{Q})$

$tan^{- 1} (\frac{P}{Q})$

$sin^{- 1} (\frac{P}{Q})$

$cos^{- 1} (- \frac{P}{Q})$

The angle made by resultant $(R)$ of the forces $P$ and $Q$ with $P$ is given by
$tan θ = \frac{Q s i n α}{P + Q c o s α}$
When $θ = 9 0^{\circ},$ then $tan 9 0^{\circ} = \infty$
$∴ P + Q cos α = 0$ or $cos α = - \frac{P}{Q}$ or
$α = cos^{- 1} (- \frac{P}{Q})$

Q. Two forces P and Q have a resultant perpendicular to P The angle between the forces is

tan−1(Q−P​)

tan−1(QP​)

sin−1(QP​)

cos−1(−QP​)