Question

.) Two forces acting at angle of 120°. The bigger force is 80N and Resultant is perpendicular to smaller one. Find the magnitude of the smaller force​

Answer 1

Answer:

Hi there, Your answer is given below

Explanation:

Let the resultant force vector  FR→  is making an angle  900  with the smaller force  F1→.  This implies that the resultant makes an angle  α=120–90=300 with the bigger vector  F2→  as shown in the figure.

From the right angle triangle ACD, defining  sinα  we get

sinα=opposite sideadjucent side=DCAD  

AD=length of F2→=F2=40N  (given)

DC=length of F1→=F1=?  (to find out)

∴sin(300)=F1F2  

⇒F1=F2×sin(300)  

⇒F1=40×12  

⇒F1=20N

Answer 2

Let the resultant force vector FR→ is making an angle 900 with the smaller force F1→. This implies that the resultant makes an angle α=120–90=300 with the bigger vector F2→

Let the resultant force vector FR→ is making an angle 900 with the smaller force F1→. This implies that the resultant makes an angle α=120–90=300 with the bigger vector F2→ As shown in the above figure.

Let the resultant force vector FR→ is making an angle 900 with the smaller force F1→. This implies that the resultant makes an angle α=120–90=300 with the bigger vector F2→ As shown in the above figure.From the right angle triangle ACD, defining sinα we get

Let the resultant force vector FR→ is making an angle 900 with the smaller force F1→. This implies that the resultant makes an angle α=120–90=300 with the bigger vector F2→ As shown in the above figure.From the right angle triangle ACD, defining sinα we getsinα=opposite sideadjucent side=DCAD

Let the resultant force vector FR→ is making an angle 900 with the smaller force F1→. This implies that the resultant makes an angle α=120–90=300 with the bigger vector F2→ As shown in the above figure.From the right angle triangle ACD, defining sinα we getsinα=opposite sideadjucent side=DCAD AD=length of F2→=F2=40N (given)

Let the resultant force vector FR→ is making an angle 900 with the smaller force F1→. This implies that the resultant makes an angle α=120–90=300 with the bigger vector F2→ As shown in the above figure.From the right angle triangle ACD, defining sinα we getsinα=opposite sideadjucent side=DCAD AD=length of F2→=F2=40N (given)DC=length of F1→=F1=? (to find out)

Let the resultant force vector FR→ is making an angle 900 with the smaller force F1→. This implies that the resultant makes an angle α=120–90=300 with the bigger vector F2→ As shown in the above figure.From the right angle triangle ACD, defining sinα we getsinα=opposite sideadjucent side=DCAD AD=length of F2→=F2=40N (given)DC=length of F1→=F1=? (to find out)∴sin(300)=F1F2

Let the resultant force vector FR→ is making an angle 900 with the smaller force F1→. This implies that the resultant makes an angle α=120–90=300 with the bigger vector F2→ As shown in the above figure.From the right angle triangle ACD, defining sinα we getsinα=opposite sideadjucent side=DCAD AD=length of F2→=F2=40N (given)DC=length of F1→=F1=? (to find out)∴sin(300)=F1F2 ⇒F1=F2×sin(300)

Let the resultant force vector FR→ is making an angle 900 with the smaller force F1→. This implies that the resultant makes an angle α=120–90=300 with the bigger vector F2→ As shown in the above figure.From the right angle triangle ACD, defining sinα we getsinα=opposite sideadjucent side=DCAD AD=length of F2→=F2=40N (given)DC=length of F1→=F1=? (to find out)∴sin(300)=F1F2 ⇒F1=F2×sin(300) ⇒F1=40×12

Let the resultant force vector FR→ is making an angle 900 with the smaller force F1→. This implies that the resultant makes an angle α=120–90=300 with the bigger vector F2→ As shown in the above figure.From the right angle triangle ACD, defining sinα we getsinα=opposite sideadjucent side=DCAD AD=length of F2→=F2=40N (given)DC=length of F1→=F1=? (to find out)∴sin(300)=F1F2 ⇒F1=F2×sin(300) ⇒F1=40×12 ⇒F1=20N

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