Determine the resultant force using the law of cosines.

Step 1: Determine the resultant force using the law of cosines.
Let the magnitudes of the two tensions be $T_1 = F$ and $T_2 = 2F$. Let the angle between the chains be $\beta$. The magnitude of the resultant force $R$ is given by:

$$R^2 = T_1^2 + T_2^2 + 2T_1 T_2 \cos\beta$$

Substitute the given tensions:

$$R^2 = F^2 + (2F)^2 + 2(F)(2F)\cos\beta$$ $$R^2 = F^2 + 4F^2 + 4F^2\cos\beta$$ $$R^2 = 5F^2 + 4F^2\cos\beta$$

Factor out $F^2$:

$$R^2 = F^2(5 + 4\cos\beta)$$

Take the square root of both sides:

$$R = \sqrt{F^2(5 + 4\cos\beta)}$$ $$R = F\sqrt{5 + 4\cos\beta}$$

This shows that the pulling force is at least $F\sqrt{5 + 4\cos\beta}$ newtons.

b) Step 1: Resolve the forces into components.
Let the stronger force, $2F$, lie along the positive x-axis. The components of the forces are:

$$\vec{T_2} = (2F, 0)$$ $$\vec{T_1} = (F\cos\beta, F\sin\beta)$$

Step 2: Calculate the components of the resultant force.
The resultant force $\vec{R} = \vec{T_1} + \vec{T_2}$ has components:

$$R_x = 2F + F\cos\beta = F(2 + \cos\beta)$$ $$R_y = F\sin\beta$$

Step 3: Determine the angle of the resultant force.
Let $\alpha$ be the angle the resultant force makes with the stronger force $\vec{T_2}$. This angle is given by:

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