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This question is incomplete, its missing an image which will be uploaded along this Answer.

Answer:

the normal component of force F_n is F((√(r²-s²)) / r)

the tangential component of force F_t is F(s/r)

Explanation:

Given the data in the image;

from the free body diagram, we write the expression for ∅

sin∅ = s/r

cos∅ = (√(r²-s²)) / r

now expression for normal component of force is;

F_n = Fcos∅

we substitute

F_n = F((√(r²-s²)) / r)

Therefore, the normal component of force F_n is F((√(r²-s²)) / r)

Also for force F_t

F_t = Fsin∅

we substitute

F_t = F(s/r)

Therefore, the tangential component of force F_t is F(s/r)

Answer:

a = 40

b = 29

Explanation:

Give a place holder for the numbers that we don't know.

Lets call the two numbers a and b.

From the given info, we can write an expression and solve it:

"one number is 11 more than another number"

a = 11 + b

from this, we know that a > b.

''three times the larger number exceeds four times the smaller number by 4"

3a = 4b + 4

Now we have 2 equations, we can use them to solve using whatever method you want.

a = 11 + b

3a = 4b + 4

I will be using matrices RREF to solve for this.

a - b = 11

3a - 4b = 4

[tex]\begin{bmatrix}1 & -1 & 11\\3 & -4 & 4 \end{bmatrix}[/tex]

[tex]\begin{bmatrix}1 & 0 & 40\\0 & 1 & 29 \end{bmatrix}[/tex]

a = 40

b = 29