Let's break down and solve the problem step-by-step.

Problem Overview:

We are asked to find the acceleration of Hachiman (whose mass is $m_H = 60.0 \, \text{kg}$) when two forces, $F_Y$ and $F_I$, are acting on him. The forces are:

• $F_Y = 125.0 \, \text{N}$ (exerted vertically),
• $F_I = 100.0 \, \text{N}$ (exerted horizontally).

We are required to find the acceleration of Hachiman in two cases:

• Part A: The forces $F_Y$ and $F_I$ are perpendicular (i.e., aligned along the $y$- and $x$-axes, respectively).
• Part B: The angle between $F_Y$ and $F_I$ is $\phi = 54^\circ$.

Part A: Forces are Perpendicular

In this case, $F_Y$ acts along the $y$-axis, and $F_I$ acts along the $x$-axis. The net force can be found using the Pythagorean theorem, since the forces are perpendicular:

$F_{\text{net}} = \sqrt{F_Y^2 + F_I^2}$

Substituting the given values:

$F_{\text{net}} = \sqrt{(125.0)^2 + (100.0)^2} = \sqrt{15625 + 10000} = \sqrt{25625}$

$F_{\text{net}} = 160.1 \, \text{N}$

Now, we can calculate the acceleration using Newton's second law, $F = ma$, where $a$ is the acceleration and $m_H = 60.0 \, \text{kg}$:

$a = \frac{F_{\text{net}}}{m_H} = \frac{160.1}{60.0} \approx 2.67 \, \text{m/s}^2$

So, the acceleration of Hachiman in Part A is approximately $2.67 \, \text{m/s}^2$.

Part B: Forces are at an Angle $\phi = 54^\circ$

In this part, $F_Y$ is at an angle of $\phi = 54^\circ$ with respect to the horizontal force $F_I$. To find the net force, we use the law of cosines:

$F_{\text{net}} = \sqrt{F_Y^2 + F_I^2 + 2 \cdot F_Y \cdot F_I \cdot \cos(\phi)}$

Substituting the given values:

$F_{\text{net}} = \sqrt{(125.0)^2 + (100.0)^2 + 2 \cdot 125.0 \cdot 100.0 \cdot \cos(54^\circ)}$

First, calculate $\cos(54^\circ)$:

$\cos(54^\circ) \approx 0.5878$

Now substitute back into the formula:

$F_{\text{net}} = \sqrt{15625 + 10000 + 2 \cdot 125.0 \cdot 100.0 \cdot 0.5878}$ $F_{\text{net}} = \sqrt{15625 + 10000 + 14720}$ $F_{\text{net}} = \sqrt{40345} \approx 200.86 \, \text{N}$

Now, we calculate the acceleration:

$a = \frac{F_{\text{net}}}{m_H} = \frac{200.86}{60.0} \approx 3.35 \, \text{m/s}^2$

So, the acceleration of Hachiman in Part B is approximately $3.35 \, \text{m/s}^2$.

Final Answers:

• Part A: $a \approx 2.67 \, \text{m/s}^2$
• Part B: $a \approx 3.35 \, \text{m/s}^2$

Would you like a more detailed explanation on any specific step, or do you have any other questions?

Related Questions:

1. How would the acceleration change if Hachiman's mass were doubled?
2. What would happen to the net force if $F_Y$ were doubled but $F_I$ stayed the same?
3. What is the direction of the net force in Part A?
4. How would the problem change if $F_I$ were at a $54^\circ$ angle instead of $F_Y$?
5. Can we solve this using vector components rather than magnitudes?

Tip:

When dealing with forces at angles, breaking them into their $x$- and $y$-components often simplifies the calculations.