Objective: Determine the acceleration of the blocks and the tension in the cord given the masses, coefficients of friction, and incline angle given:
Part a: Calculate the acceleration of the blocks
Step | Formula Derivation | Reasoning |
---|---|---|
1 | F_{\text{gravity, parallel A}} = m_A g \sin(\theta) | Parallel component of gravitational force for A. |
2 | F_{\text{gravity, parallel B}} = m_B g \sin(\theta) | Parallel component of gravitational force for B. |
3 | F_{\text{friction A}} = \mu_A m_A g \cos(\theta) | Frictional force on A. |
4 | F_{\text{friction B}} = \mu_B m_B g \cos(\theta) | Frictional force on B. |
5 | F_{\text{net A}} = F_{\text{gravity, parallel A}} – F_{\text{friction A}} | Net force on A. |
6 | F_{\text{net B}} = F_{\text{gravity, parallel B}} – F_{\text{friction B}} | Net force on B. |
7 | F_{\text{net}} = F_{\text{net B}} – F_{\text{net A}} | Total net force on the system. |
8 | a = \frac{F_{\text{net}}}{m_A + m_B} | Acceleration of the system. |
Plug in the given values:
Step | Formula Derivation | Reasoning |
---|---|---|
9 | a = \frac{(m_B g \sin(\theta) – \mu_B m_B g \cos(\theta)) – (m_A g \sin(\theta) – \mu_A m_A g \cos(\theta))}{m_A + m_B} | Substitute the net forces from steps 5 and 6. |
10 | a = \frac{(5 \cdot 9.8 \cdot \sin(32^\circ) – 0.30 \cdot 5 \cdot 9.8 \cdot \cos(32^\circ)) – (5 \cdot 9.8 \cdot \sin(32^\circ) – 0.20 \cdot 5 \cdot 9.8 \cdot \cos(32^\circ))}{5 + 5} | Substitute given values. |
11 | a = \frac{(5 \cdot 9.8 \cdot (\sin(32^\circ) – 0.30 \cdot \cos(32^\circ))) – (5 \cdot 9.8 \cdot (\sin(32^\circ) – 0.20 \cdot \cos(32^\circ)))}{10} | Simplify the expression. |
12 | a = \frac{5 \cdot 9.8 \cdot (0.10 \cdot \cos(32^\circ))}{10} | Combine like terms. |
13 | a = \frac{9.8 \cdot (0.10 \cdot \cos(32^\circ))}{2} | Simplify further. |
Part b: Calculate the tension in the cord
Step | Formula Derivation | Reasoning |
---|---|---|
1 | T = m_A \cdot a + F_{\text{friction A}} | Tension equals the force to accelerate block A plus frictional force on A. |
2 | F_{\text{friction A}} = \mu_A \cdot m_A \cdot g \cdot \cos(\theta) | Frictional force opposing the motion of block A. |
3 | T = m_A \cdot a + \mu_A \cdot m_A \cdot g \cdot \cos(\theta) | Substitute the frictional force into the tension formula. |
4 | T = m_A \cdot \left( \frac{F_{\text{net}}}{m_A + m_B} \right) + \mu_A \cdot m_A \cdot g \cdot \cos(\theta) | Substitute the expression for �a from the acceleration calculation. |
5 | T = 5 \cdot \left( \frac{(5 \cdot 9.8 \cdot \sin(32^\circ) – 0.30 \cdot 5 \cdot 9.8 \cdot \cos(32^\circ)) – (5 \cdot 9.8 \cdot \sin(32^\circ) – 0.20 \cdot 5 \cdot 9.8 \cdot \cos(32^\circ))}{10} \right) + 0.20 \cdot 5 \cdot 9.8 \cdot \cos(32^\circ) | Insert given values for masses, coefficients of friction, gravitational acceleration, and angle. |
6 | T = 5 \cdot \left( \frac{5 \cdot 9.8 \cdot (0.10 \cdot \cos(32^\circ))}{10} \right) + 0.20 \cdot 5 \cdot 9.8 \cdot \cos(32^\circ) | Simplify the expression for the net force component. |
7 | T = \frac{5 \cdot 9.8 \cdot (0.10 \cdot \cos(32^\circ))}{2} + 0.20 \cdot 5 \cdot 9.8 \cdot \cos(32^\circ) | Further simplify the tension formula. |
8 | T = \frac{5 \cdot 9.8 \cdot 0.10 \cdot \cos(32^\circ)}{2} + 0.20 \cdot 5 \cdot 9.8 \cdot \cos(32^\circ) | Combine like terms for the final tension calculation. |
9 | T \approx 17.66 , \text{N} | Calculate the numeric value for tension. |
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Block m1 is stacked on top of block m2. Block m2 is connected by a light cord to block m3, which is pulled along a frictionless surface with a force F as shown in the diagram above. Block m1 is accelerated at the same rate as block m2 because of the frictional forces between the two blocks. If all three blocks have the same mass m, what is the minimum coefficient of static friction between block m1 and block m2?
On a harsh winter day, a 1500 kg vehicle takes a circular banked exit ramp (radius R = 150 m; banking angle of 10 degrees) at a speed of 30 mph, since the speed limit is 35 mph. However, the exit ramp is completely iced up (= frictionless). To make matters worse, a wind is blowing parallel to the ramp in a downward direction. The wind exerts a force of 3000 N. Under these conditions, can the driver continue to follow a safe horizontal circle on the exit ramp and stay below the speed limit? To convert mph into m/s use 1 mi = 1607 m and 1 hr is 3600 s.
A skydiver reaches a terminal velocity of 55.0 m/s. At terminal velocity, the skydiver no longer accelerates. The mass of the skydiver and her equipment is 87.0 kg. What is the force of friction acting on her?
What condition(s) are necessary for static equilibrium?
A person whose weight is 4.92 × 102 N is being pulled up vertically by a rope from the bottom of a cave that is 35.2 m deep. The maximum tension that the rope can withstand without breaking is 592 N. What is the shortest time, starting from rest, in which the person can be brought out of the cave?
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Kinematics | Forces |
---|---|
\Delta x = v_i t + \frac{1}{2} at^2 | F = ma |
v = v_i + at | F_g = \frac{G m_1m_2}{r^2} |
a = \frac{\Delta v}{\Delta t} | f = \mu N |
R = \frac{v_i^2 \sin(2\theta)}{g} |
Circular Motion | Energy |
---|---|
F_c = \frac{mv^2}{r} | KE = \frac{1}{2} mv^2 |
a_c = \frac{v^2}{r} | PE = mgh |
KE_i + PE_i = KE_f + PE_f |
Momentum | Torque and Rotations |
---|---|
p = m v | \tau = r \cdot F \cdot \sin(\theta) |
J = \Delta p | I = \sum mr^2 |
p_i = p_f | L = I \cdot \omega |
Simple Harmonic Motion |
---|
F = -k x |
T = 2\pi \sqrt{\frac{l}{g}} |
T = 2\pi \sqrt{\frac{m}{k}} |
Constant | Description |
---|---|
g | Acceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface |
G | Universal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2 |
\mu_k and \mu_s | Coefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion. |
k | Spring constant, in \text{N/m} |
M_E = 5.972 \times 10^{24} , \text{kg} | Mass of the Earth |
M_M = 7.348 \times 10^{22} , \text{kg} | Mass of the Moon |
M_M = 1.989 \times 10^{30} , \text{kg} | Mass of the Sun |
Variable | SI Unit |
---|---|
s (Displacement) | \text{meters (m)} |
v (Velocity) | \text{meters per second (m/s)} |
a (Acceleration) | \text{meters per second squared (m/s}^2\text{)} |
t (Time) | \text{seconds (s)} |
m (Mass) | \text{kilograms (kg)} |
Variable | Derived SI Unit |
---|---|
F (Force) | \text{newtons (N)} |
E, PE, KE (Energy, Potential Energy, Kinetic Energy) | \text{joules (J)} |
P (Power) | \text{watts (W)} |
p (Momentum) | \text{kilogram meters per second (kgm/s)} |
\omega (Angular Velocity) | \text{radians per second (rad/s)} |
\tau (Torque) | \text{newton meters (Nm)} |
I (Moment of Inertia) | \text{kilogram meter squared (kgm}^2\text{)} |
f (Frequency) | \text{hertz (Hz)} |
General Metric Conversion Chart
Example of using unit analysis: Convert 5 kilometers to millimeters.
Start with the given measurement: \text{5 km}
Use the conversion factors for kilometers to meters and meters to millimeters: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}
Perform the multiplication: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}
Simplify to get the final answer: \boxed{5 \times 10^6 \, \text{mm}}
Prefix | Symbol | Power of Ten | Equivalent |
---|---|---|---|
Pico- | p | 10^{-12} | 0.000000000001 |
Nano- | n | 10^{-9} | 0.000000001 |
Micro- | µ | 10^{-6} | 0.000001 |
Milli- | m | 10^{-3} | 0.001 |
Centi- | c | 10^{-2} | 0.01 |
Deci- | d | 10^{-1} | 0.1 |
(Base unit) | – | 10^{0} | 1 |
Deca- or Deka- | da | 10^{1} | 10 |
Hecto- | h | 10^{2} | 100 |
Kilo- | k | 10^{3} | 1,000 |
Mega- | M | 10^{6} | 1,000,000 |
Giga- | G | 10^{9} | 1,000,000,000 |
Tera- | T | 10^{12} | 1,000,000,000,000 |
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