The sum of two forces at a point is $16 N$ . If the resultant is normal to the smaller force and has a value $8 N$ , find the two forces.

16 N, 0 N

No worries! We‘ve got your back. Try BYJU‘S free classes today!

10 N, 6 N

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

60 N, 10 N

No worries! We‘ve got your back. Try BYJU‘S free classes today!

30 N, 7 N

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $10 N, 6 N$
Given,
$F_{1}^{2} = F_{2}^{2} + F_{R}^{2}$
$⟹ F_{1}^{2} - F_{2}^{2} = F_{R}^{2}$
$⟹ (F_{1} + F_{2}) (F_{1} - F_{2}) = 64$
$⟹ F_{1} - F_{2} = \frac{64}{16} = 4$
Hence,
$F_{1} - F_{2} = 4$
$F_{1} + F_{2} = 16$
Solving,
$F_{1} = 10, F_{2} = 6$

Suggest Corrections

Similar questions

The sum of two forces acting at a point is 16 N. If the resultant force is 8 N and its direction is perpendicular to minimum force then the forces are

Q. The sum of the magnitudes of two forces acting at a point is

16 N

. The resultant of these forces is perpendicular to the smaller force and has a magnitude of

8 N

. If the smaller force is of magnitude

x

, then the value of

x

Q. Two forces are such that the sum of their magnitudes is

18 N

and their resultant is perpendicular to the smaller force and magnitude of resultant is

12 N

. Then the magnitude of the forces are

Q. The sum of magnitudes of two forces acting at a point is

16 N

. If the resultant force is

8 N

and its direction is perpendicular to smaller force, then the forces are:

Two forces are such that the sum of their magnitudes is 18 N and their resultant is perpendicular to the smaller force and magnitude of resultant is 12 N. Then the magnitudes of the forces are

Join BYJU'S Learning Program

The sum of two forces at a point is 16N. If the resultant is normal to the smaller force and has a value 8N, find the two forces.

The correct option is B 10N,6NGiven, F21=F22+F2R⟹F21−F22=F2R⟹(F1+F2)(F1−F2)=64⟹F1−F2=6416=4Hence,F1−F2=4F1+F2=16Solving,F1=10,F2=6

The sum of two forces at a point is $16 N$ . If the resultant is normal to the smaller force and has a value $8 N$ , find the two forces.

The correct option is B $10 N, 6 N$
Given,
$F_{1}^{2} = F_{2}^{2} + F_{R}^{2}$
$⟹ F_{1}^{2} - F_{2}^{2} = F_{R}^{2}$
$⟹ (F_{1} + F_{2}) (F_{1} - F_{2}) = 64$
$⟹ F_{1} - F_{2} = \frac{64}{16} = 4$
Hence,
$F_{1} - F_{2} = 4$
$F_{1} + F_{2} = 16$
Solving,
$F_{1} = 10, F_{2} = 6$