Since the body is initially at equilibrium under the action of the four forces, the net force on the body is zero.
This gives us the equation for equilibrium: \[ F_1 + F_2 + F_3 + F_4 = 0 \] When \( F_1 \) is removed, the remaining forces acting on the body are \( F_2, F_3, \) and \( F_4 \).
The net force on the body is now: \[ F_{{net}} = F_2 + F_3 + F_4 \] From the equilibrium condition, we know that \( F_1 = -(F_2 + F_3 + F_4) \).
Therefore, when \( F_1 \) is removed, the remaining net force is: \[ F_{{net}} = F_1 \] Thus, the acceleration \( a \) of the body is given by Newton's second law: \[ a = \frac{F_{{net}}}{M} = \frac{F_1}{M} \]
Therefore, the body moves with an acceleration of \( \frac{F_1}{M} \) after \( F_1 \) is removed.
Thus, the correct answer is: \[ \boxed{\frac{F_1}{M}} \]
\[ f(x) = \begin{cases} x\left( \frac{\pi}{2} + x \right), & \text{if } x \geq 0 \\ x\left( \frac{\pi}{2} - x \right), & \text{if } x < 0 \end{cases} \]
Then \( f'(-4) \) is equal to:If \( f'(x) = 4x\cos^2(x) \sin\left(\frac{x}{4}\right) \), then \( \lim_{x \to 0} \frac{f(\pi + x) - f(\pi)}{x} \) is equal to:
Let \( f(x) = \frac{x^2 + 40}{7x} \), \( x \neq 0 \), \( x \in [4,5] \). The value of \( c \) in \( [4,5] \) at which \( f'(c) = -\frac{1}{7} \) is equal to:
The general solution of the differential equation \( \frac{dy}{dx} = xy - 2x - 2y + 4 \) is:
\[ \int \frac{4x \cos \left( \sqrt{4x^2 + 7} \right)}{\sqrt{4x^2 + 7}} \, dx \]