Let the force from the horse be $\vec{F}_H$ and the force from the water be $\vec{F}_W$. The barge's acceleration is $\vec{a} = a\hat{i}$. We apply Newton's Second Law, $\sum \vec{F} = m\vec{a}$.

$$ \vec{F}_H + \vec{F}_W = m\vec{a} $$

Solving for the force from the water gives:

$$ \vec{F}_W = m\vec{a} - \vec{F}_H $$

We can find the components of $\vec{F}_W$ by resolving the vectors into x and y components.

$$ \vec{F}_H = (F_H \cos\theta)\hat{i} + (F_H \sin\theta)\hat{j} $$ $$ m\vec{a} = (ma)\hat{i} $$

The components of $\vec{F}_W = F_{W,x}\hat{i} + F_{W,y}\hat{j}$ are therefore:

$$ F_{W,x} = ma - F_H \cos\theta $$ $$ F_{W,y} = 0 - F_H \sin\theta = -F_H \sin\theta $$

The magnitude $F_W$ is found using the Pythagorean theorem:

$$ F_W = \sqrt{F_{W,x}^2 + F_{W,y}^2} = \sqrt{(ma - F_H \cos\theta)^2 + (-F_H \sin\theta)^2} $$ $$ F_W = \sqrt{m^2a^2 - 2maF_H\cos\theta + F_H^2\cos^2\theta + F_H^2\sin^2\theta} $$ $$ F_W = \sqrt{(ma)^2 + F_H^2 - 2maF_H\cos\theta} $$

The direction $\phi$ is found from the ratio of the components:

$$ \tan\phi = \frac{F_{W,y}}{F_{W,x}} = \frac{-F_H \sin\theta}{ma - F_H \cos\theta} $$