Part (a): Drawing a Free Body Diagram (FBD)
For the purposes of this explanation, drawing of the FBD is described:
– The ladder rests against a wall with length L and weight W = 50 \, \text{N} .
– F_N represents the normal force exerted by the wall on the ladder, acting horizontally at the top of the ladder.
– F_f is the frictional force at the base of the ladder, which opposes the sliding movement, acting horizontally towards the wall.
– N stands for the normal force exerted by the ground on the ladder, vertically upwards.
– W is the weight of the ladder acting downwards from its center of mass.
Part (b): Finding the Minimum Angle \theta_{\text{min}} so the Ladder Does Not Slip
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \text{Sum of horizontal forces: } F_N = F_f | The normal force F_N exerted by the wall is balanced by the frictional force F_f at the base, since there is no other horizontal motion. |
| 2 | \text{Sum of vertical forces: } N = W | The normal force N from the ground balances the gravitational force W because there is no vertical motion. |
| 3 | F_f = \mu N | The frictional force F_f can be expressed as the product of the coefficient of static friction \mu and the normal force N . |
| 4 | \text{Torque about point at base of ladder: } \\ W \frac{L}{2} \cos(\theta) = F_N L \sin(\theta) | Taking torque about the base, counterclockwise torques due to the wall’s normal force F_N should balance the clockwise torque due to the weight W . The distance from the base to the CM is \frac{L}{2} . |
| 5 | \frac{W}{2} = F_N \tan(\theta) | Further simplification. |
| 6 | \frac{W}{2} = \mu W \tan(\theta) | Replace F_N with \mu N as solved for in step 3, since F_N = F_f as described in step 1. |
| 7 | \tan(\theta) = \frac{1}{2\mu} | Simplify further by canceling out W and isolating \theta . |
| 8 | \theta = \tan^{-1}(\frac{1}{2\mu}) | Finally, take the inverse tan to find \theta . |
| 9 | \theta = \tan^{-1}(\frac{1}{2 \times .4}) | Substitute \mu = 0.4 into the equation derived. |
| 10 | \theta = \tan^{-1}(0.8) \approx 51.34^\circ | This is the minimum angle to ensure the ladder does not slip. Calculate \theta in degrees. |
| 11 | \boxed{\theta_{\text{min}} \approx 51.34^\circ} | Final answer. |
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An object weighing 120 N is set on a rigid beam of negligible mass at a distance of 3 m from a pivot, as shown above. A vertical force is to be applied to the other end of the beam a distance of 4 m from the pivot to keep the beam at rest and horizontal. What is the magnitude F of the force required?
A 150-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope.
What constant force must be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.500 rev/s in 2.00 s?
Note: I_\text{disk} = \frac{1}{2}mr^2
An ice skater performs a pirouette (a fast spin) by pulling in his outstretched arms close to his body. What happens to his angular momentum about the axis of rotation?
A rod is initially at rest on a rough horizontal surface. Three forces are exerted on the rod with the magnitudes and directions shown in the figure. The force exerted in the center of the rod is an equidistant 0.5 m from both ends of the rod. If friction between the rod and the table prevents the rod from rotating, what is the magnitude of the torque exerted on the rod about its center from frictional forces?
Suppose a solid uniform sphere of mass M and radius R rolls without slipping down an inclined plane starting from rest. The angular velocity of the sphere at the bottom of the incline depends on
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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} | |
Nano- | n | 10^{-9} | |
Micro- | µ | 10^{-6} | |
Milli- | m | 10^{-3} | |
Centi- | c | 10^{-2} | |
Deci- | d | 10^{-1} | |
(Base unit) | – | 10^{0} | |
Deca- or Deka- | da | 10^{1} | |
Hecto- | h | 10^{2} | |
Kilo- | k | 10^{3} | |
Mega- | M | 10^{6} | |
Giga- | G | 10^{9} | |
Tera- | T | 10^{12} |
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