Question

the sum of two forces at a point 16 N. if the resultant is normal to the smaller force have value 8 N. find two forces.​

Answer 1

Answer:

The forces are 6 N and 10 N

Explanation:

given that,

the sum of two forces at a point is 16 N

let the two forces be A and B

so,

here given,

A + B = 16 .....(1)

also given that,

the resultant is normal to the smaller force have value 8 N,

so,

here,

sum of vectors = 8 N

i.e

A^--> + B^-> = 8 N

so,

Resultant(R) of forces = 8 N

now,

ACCORDING TO THE FIGURE

R² = A² + B² + 2(AB)cosθ

putting the values,

8² = A² + B² + 2AB cosθ

64 = A² + B² + 2AB cosθ ...(2)

now,

by the direction formulae

tanα = Bsinθ/(A + Bcosθ)

here,

α = 90 [according to the figure]

putting the values,

tan 90 = Bsinθ/(A + Bcosθ)

ж = Bsinθ/(A + Bcosθ)

ж = infinity

so,

A + Bcosθ = 0

Bcosθ = -A

putting the value of Bcosθ on (2)

64 = A² + B² + 2ABcosθ

64 = A² + B² + 2A(-A)

64 = A² + B² - 2A²

64 = B² - A²

64 = (B + A) (B - A)

from (1)

B + A = 16

putting the value

64 = (B + A) (B - A)

64 = 16(B - A)

B - A = 64/16

B - A = 4. ..(3)

now,

we have

B + A = 16 ..(1)

B - A = 4 ...(2)

adding both equation,

B + A + B - A = 16 + 4

2B = 20

B = 20/2

B = 10

now,

putting the value of B on (2)

B - A = 4

10 - A = 4

-A = 4 - 10

-A = -6

A = 6

so,

________________

The forces are

6 N and 10 N

_______________

now,

for finding the angles between the forces

64 = A² + B² + 2ABcosθ

putting the values

64 = 6² + 10² + 2(6)(10) cosθ

64 = 36 + 100 + 120 cos θ

120cosθ + 136 = 64

120cosθ = 64 - 136

120cosθ = 72

cos θ = 72/120

cos θ = 3/5

so,

θ = 37 °

Angle between the forces = 37°

Answer 2

Answer:

The forces are

The forces are6 N and 10 N

Explanation:

Refer to attachment