The problem seems to involve forces exerted by multiple players on a box, which appears to be on a frictionless surface, and involves determining the force components exerted by Player 3 and then calculating the box's acceleration and mass after the rope from Player 3 breaks.

Let’s break this down into two main parts:

Part 1: Determining the Force Components Exerted by Player 3

Given:

• Force exerted by Player 1: $F_1 = 300 \, \text{N}$
• Angle $\theta_1 = -60^\circ$ (below the x-axis)

To determine the force components of Player 3 ($F_{3x}$ and $F_{3y}$), we need to know more information about Player 2's force (magnitude and direction) and how all the forces add up to balance the system.

Let’s denote Player 2's force by $F_2$, acting at some angle $\theta_2$. Since this part asks for the components of Player 3’s force, the sum of the forces from Players 1, 2, and 3 should result in equilibrium (no acceleration), assuming this system is initially in equilibrium.

The total force in the x-direction and y-direction is zero: $\sum F_x = F_{1x} + F_{2x} + F_{3x} = 0$ $\sum F_y = F_{1y} + F_{2y} + F_{3y} = 0$

We can calculate the components of $F_1$ as: $F_{1x} = F_1 \cos(\theta_1) = 300 \times \cos(-60^\circ) = 300 \times 0.5 = 150 \, \text{N}$ $F_{1y} = F_1 \sin(\theta_1) = 300 \times \sin(-60^\circ) = 300 \times (-0.866) = -259.8 \, \text{N}$

To proceed, we need the magnitude and direction of the force exerted by Player 2.

Part 2: After Player 3's Rope Breaks

After Player 3's rope breaks, Player 2 adjusts by pulling with a force of $F'_2 = 350 \, \text{N}$, but still at the same angle $\theta_2$.

Next, we need to determine the angle $\theta$ and mass $m$ of the box. When Player 3's force is removed, the net force from Player 1 and the adjusted force from Player 2 will cause the box to accelerate.

The total force in this case is given by: $F_{\text{net}} = \sqrt{F_{1x}^2 + F_{1y}^2} + \sqrt{F_{2x}^2 + F_{2y}^2}$ Given that the magnitude of acceleration is $a = 10 \, \text{m/s}^2$, we can use Newton’s second law to find the mass $m$: $F_{\text{net}} = m \cdot a$ From this, we can solve for $m$: $m = \frac{F_{\text{net}}}{a}$

To complete the calculation, I need a bit more information:

• The magnitude and direction of the force exerted by Player 2 in the initial part of the problem.
• Confirmation on whether the system was in equilibrium before Player 3's rope broke.

Would you like to clarify this or upload a figure for reference?