Question

Two forces of magnitude 8 N and 15 N respectively
act at a point. If the resultant force is 17 N, the
angle between the forces has to be
(1) 60°
(2) 45°
(3) 90°
(4) 30°
​

Answer 1

Answer:

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Answer 2

Answer :

option (3) 90°

Explanation :

  Let θ be the angle between the two forces.

Given,

• Two forces of magnitude 8 N and 15 N respectively  act at a point.

• The resultant force is 17 N

To find,

• The  angle between the forces, θ = ?

      $Let \ P=8 \ N \ and \ Q = 15 \ N \\ and \ R=17 \ N$

$The \ resultant \ of \ the \ two \ vectors \ \vec{P} \ and \ \vec{Q} \ is \ given \ by$

                  $\boxed{\bf R=\sqrt{P^2+Q^2+2PQcos\theta}}$

           

        $\longrightarrow \ \ \ \ 17=\sqrt{8^2+15^2+2(8)(15)cos\theta} \\\\ \longrightarrow \ \ \ \ 17^2=8^2+15^2+(16)(15)cos\theta \\\\ \longrightarrow \ \ \ \ 289=64+225+240cos\theta \\\\ \longrightarrow \ \ \ \ 289=289+240cos\theta \\\\ \longrightarrow \ \ \ \ 289-289=240cos\theta \\\\ \longrightarrow \ \ \ \ 0=240cos\theta \\\\ \longrightarrow \ \ \ \ cos\theta=0 \\\\ \longrightarrow \ \ \ \ cos\theta=cos90^0 \\\\ \longrightarrow \ \ \ \ \boxed{\theta=90^0}$

∴ The angle between the forces is 90°

 ______________________________

⇒   MORE INFORMATION :

• Resultant is the vector sum of two or more vectors.
• Direction of resultant,

            Let α be the angle made by the resultant R with P

     From ΔOCD, (refer the attachment)

                    tan α = CD/OD

                    tan α = CD/(OA+AD)

                    $tan\alpha =\frac{Q \ sin\theta}{P+Q \ cos \theta} \\\\ \bf \alpha =tan^{-1}(\frac{Q \ sin\theta}{P+Q \ cos \theta})$