.) 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
Answers
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
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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