ABC is a right_angled triangle at $$ B, A B=6 \mathrm{~cm}, B C=8 \mathrm{~cm} $$, a force $$ \vec{F} $$ acts in the plane of the triangle such that $$ M_{A}=M_{B}=60 \mathrm{~N} . \mathrm{cm} M_{C}=-60 \mathrm{~N} . \mathrm{cm} $$ ,then $$ \|\vec{F}\|=\cdots N $$

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Given the right-angled triangle $$ ABC $$ with the right angle at $$ B $$, we have:

- $$ AB = 6 $$ cm
- $$ BC = 8 $$ cm
- $$ AC = \sqrt{6^2 + 8^2} = 10 $$ cm

A force $$ \vec{F} $$ acts in the plane of the triangle, and the moments about the points $$ A $$, $$ B $$, and $$ C $$ are given as:
- $$ M_A = 60 $$ N·cm
- $$ M_B = 60 $$ N·cm
- $$ M_C = -60 $$ N·cm

Let's set up a coordinate system with $$ B $$ at the origin:
- $$ B = (0, 0) $$
- $$ A = (0, 6) $$
- $$ C = (8, 0) $$

Assume $$ \vec{F} $$ has components $$ F_x $$ and $$ F_y $$, and it acts at a point $$ P = (x, y) $$. The moments about points $$ A $$, $$ B $$, and $$ C $$ can be expressed as:

1. $$ M_A = x F_y - (y - 6) F_x = 60 $$
2. $$ M_B = x F_y - y F_x = 60 $$
3. $$ M_C = (x - 8) F_y - y F_x = -60 $$

Subtracting the second equation from the first:
$$ 6 F_x = 0 \implies F_x = 0 $$

From the second equation:
$$ x F_y = 60 $$

From the third equation:
$$ x F_y - 8 F_y = -60 $$
Since $$ x F_y = 60 $$:
$$ 60 - 8 F_y = -60 \implies F_y = 15 \text{ N} $$

Thus, the force $$ \vec{F} $$ has components $$ F_x = 0 $$ and $$ F_y = 15 $$ N. The magnitude of the force is:
$$ \| \vec{F} \| = \sqrt{F_x^2 + F_y^2} = 15 \text{ N} $$

**Answer:**  
The magnitude of the force F is fifteen newtons
The magnitude of the force $$ \vec{F} $$ is 15 N.

ABC is a right_angled triangle at , a force acts
in the plane of the triangle such that
,then

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