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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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1. An FBD with 5 forces must be drawn or described for this question.
2. \theta_{\text{min}} \approx 51.34^\circ

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KinematicsForces
\Delta x = v_i t + \frac{1}{2} at^2F = ma
v = v_i + atF_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 MotionEnergy
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
MomentumTorque and Rotations
p = m v\tau = r \cdot F \cdot \sin(\theta)
J = \Delta pI = \sum mr^2
p_i = p_fL = I \cdot \omega
Simple Harmonic Motion
F = -k x
T = 2\pi \sqrt{\frac{l}{g}}
T = 2\pi \sqrt{\frac{m}{k}}
ConstantDescription
gAcceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface
GUniversal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2
\mu_k and \mu_sCoefficients 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.
kSpring 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
VariableSI 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)}
VariableDerived 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)}
\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.

2. 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}}

3. 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}

4. 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}

1. Some answers may be slightly off by 1% depending on rounding, etc.
2. Answers will use different values of gravity. Some answers use 9.81 m/s2, and other 10 m/s2 for calculations.
3. Variables are sometimes written differently from class to class. For example, sometime initial velocity v_i is written as u ; sometimes \Delta x is written as s .
4. Bookmark questions that you can’t solve so you can come back to them later.
5. Always get help if you can’t figure out a problem. The sooner you can get it cleared up the better chances of you not getting it wrong on a test!

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