Player 1 exerts a force Fi equal to 300 N at an angle 0, equal to -60.0° with respect to the +x direction. Player 2 exerts a
respect to the +x direction, as shown in the figure.
The view in the figure is from above. Friction and gravity may be ignored in this problem.
0.
F
Determine the force F3 exerted by player 3. State your answer by giving the x and y components F3x and F3y, respectively.
F3x =
N
F3y =
Player 3's rope breaks, and player 2 adjusts by pulling with a force of magnitude F' equal to 350 N at the same angle as before.
Defining angles above the x axis as positive and those below as negative, at what angle 0 is the acceleration of the box relative to the +x direction?
0 =
N
The magnitude of the acceleration is measured to be 10,0 m/?.
What is the mass m of the box?
m

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?

Player 1 exerts a force Fi equal to 300 N at an angle of -60.0° with respect to the +x direction. Player 2 exerts a force at a different angle, as shown in the figure. The system is in equilibrium, and the view is from above, with friction and gravity ignored. Determine the force F3 exerted by Player 3, giving the x and y components (F3x and F3y, respectively). After Player 3's rope breaks, Player 2 adjusts by pulling with a force of 350 N at the same angle. At what angle θ does the box accelerate relative to the +x direction, and what is the mass of the box given that the acceleration is 10 m/s²?

This problem involves finding the force components exerted by Player 3 on a box in equilibrium and determining the acceleration and mass of the box after Player 3's rope breaks. It covers concepts such as vector components, force equilibrium, and Newton's second law. Suitable for college-level students in Physics or Engineering.

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