A Mole Is Digging A Tunnel From The Base Of A Tree
A Mole Is Digging A Tunnel From The Base Of A Tree It Digs 264 M No
A mole is digging a tunnel from the base of a tree. It digs 2.64 m north then 4.73 m at 43.8º east of north. How far is the mole relative to the starting point?
A ball rolls off the lab benchtop 0.92 m from the ground and strike the ground 0.538 m from the lab bench. What is the velocity of the ball when it leaves the benchtop?
A 5.1 kg mass is hanging with the support of cables where the angles a = 25.4º and b = 39.2º. Determine the tensions of T1.
Paper For Above instruction
Introduction
Understanding the principles of vector displacement, projectile motion, and static equilibrium is essential in solving various physics problems. These problems involve calculating distances, velocities, and tension forces using trigonometry, kinematic equations, and equilibrium conditions. In this paper, we analyze three different scenarios: the displacement of a mole digging a tunnel, the velocity of a ball leaving a benchtop, and the tension in supporting cables holding a mass.
Problem 1: Displacement of the Mole
The first problem involves calculating the net displacement of a mole that digs a tunnel with two segments. The initial segment is 2.64 meters north, and the second is 4.73 meters at an angle of 43.8° east of north. To find the total displacement relative to the starting point, it is necessary to resolve these vectors into their components and then compute the resultant vector's magnitude.
The first vector, \( \vec{A} \), points due north with a magnitude of 2.64 m. Its components are:
\[
A_x = 0 \quad \text{(since it points purely north)} \\
A_y = 2.64 \quad \text{(north component)}
\]
The second vector, \( \vec{B} \), has a magnitude of 4.73 m at 43.8° east of north. Its components along the north-south axis and east-west axis are:
\[
B_x = 4.73 \sin 43.8^\circ \\
B_y = 4.73 \cos 43.8^\circ
\]
Calculating these:
\[
B_x \approx 4.73 \times 0.690 \approx 3.263 \text{ m} \\
B_y \approx 4.73 \times 0.723 \approx 3.422 \text{ m}
\]
Total components:
\[
R_x = 0 + 3.263 \approx 3.263 \text{ m} \\
R_y = 2.64 + 3.422 \approx 6.062 \text{ m}
\]
The magnitude of the resultant displacement:
\[
R = \sqrt{R_x^2 + R_y^2} \approx \sqrt{(3.263)^2 + (6.062)^2} \approx \sqrt{10.65 + 36.78} \approx \sqrt{47.43} \approx 6.89 \text{ m}
\]
Thus, the mole is approximately 6.89 meters from the starting point.
Problem 2: Velocity of the Ball
The second problem involves projectile motion, where a ball rolls off a bench 0.92 meters high and lands 0.538 meters from the edge. The goal is to determine the initial velocity when it leaves the bench.
First, find the time of flight, \( t \), using vertical displacement:
\[
h = \frac{1}{2} g t^2
\]
\[
t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 0.92}{9.8}} \approx \sqrt{\frac{1.84}{9.8}} \approx \sqrt{0.188} \approx 0.434 \text{ s}
\]
Next, calculate the horizontal velocity \( v_x \):
\[
v_x = \frac{\Delta x}{t} = \frac{0.538}{0.434} \approx 1.24 \text{ m/s}
\]
Since the initial vertical velocity is zero at the moment of leaving the bench, the total initial velocity is primarily horizontal:
\[
v = v_x \approx 1.24 \text{ m/s}
\]
Note: If the ball has any vertical initial velocity component, the problem would need more information to resolve this. Based on the given data, the velocity of the ball as it leaves the bench is approximately 1.24 m/s.
Problem 3: Tensions in Cables Supporting a Mass
The final problem involves a mass of 5.1 kg supported by two cables at angles \( a = 25.4^\circ \) and \( b = 39.2^\circ \). Assuming the system is in static equilibrium, the tension forces \( T_1 \) and \( T_2 \) in the cables are related by the conditions:
\[
\sum F_x = 0, \quad \sum F_y = 0
\]
Decompose the tensions into components:
\[
T_1 \text{ (angle } a): \quad T_1_x = T_1 \cos a, \quad T_1_y = T_1 \sin a
\]
\[
T_2 \text{ (angle } b): \quad T_2_x = T_2 \cos b, \quad T_2_y = T_2 \sin b
\]
Since the tension components balance the weight:
\[
T_1 \sin a + T_2 \sin b = mg
\]
\[
T_1 \cos a = T_2 \cos b
\]
From the second equation:
\[
T_2 = T_1 \frac{\cos a}{\cos b}
\]
Substitute into the first:
\[
T_1 \sin a + T_1 \frac{\cos a}{\cos b} \sin b = mg
\]
\[
T_1 \left( \sin a + \frac{\cos a \sin b}{\cos b} \right) = mg
\]
Calculate numerical values:
\[
mg = 5.1 \times 9.8 \approx 49.98 \, \text{N}
\]
\[
\sin 25.4^\circ \approx 0.429, \quad \cos 25.4^\circ \approx 0.903
\]
\[
\sin 39.2^\circ \approx 0.632, \quad \cos 39.2^\circ \approx 0.776
\]
Now:
\[
T_1 \left( 0.429 + \frac{0.903 \times 0.632}{0.776} \right) = 49.98
\]
Compute the numerator:
\[
0.903 \times 0.632 \approx 0.571
\]
Divide:
\[
0.571 / 0.776 \approx 0.736
\]
Sum:
\[
0.429 + 0.736 = 1.165
\]
Finally, solve for \( T_1 \):
\[
T_1 = \frac{49.98}{1.165} \approx 42.9\, \text{N}
\]
and
\[
T_2 = T_1 \times \frac{0.903}{0.776} \approx 42.9 \times 1.164 \approx 50.0\, \text{N}
\]
Conclusion: The tension in cable T1 is approximately 42.9 N, and in T2 is approximately 50.0 N.
Conclusion
These problems illustrate the application of vector addition, kinematic equations, and static equilibrium principles. Precise resolution into components enables accurate determination of distances, velocities, and forces. Solving such problems enhances comprehension of fundamental physics concepts essential for various scientific and engineering applications.
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