## Supercharge UBQ with

0 attempts

0% avg

UBQ Credits

Verfied Explanation 0 likes
0
Step Derivation/Formula Reasoning
1 F_{\text{gravity}} = mg Calculate the gravitational force acting on the crate. This uses the mass of the crate m = 25.0 \, \text{kg} and the acceleration due to gravity g = 9.8 \, \text{m/s}^2.
2 F_{\text{gravity}} = 25.0 \times 9.8 = 245 \, \text{N} Performing the calculation gives the gravitational force acting on the crate.
3 F_{\text{gravity, parallel}} = F_{\text{gravity}} \sin(\theta) Determine the component of the gravitational force that acts parallel to the slope. Here, \theta = 38.0^\circ.
4 F_{\text{gravity, parallel}} = 245 \sin(38.0^\circ) = 150.8 \, \text{N} Calculation of the parallel component of gravitational force using the angle’s sine value.
4.5 F_{\text{gravity, parallel}} = F_{\text{friction needed}} Answer – \boxed{150.8 \, \text{N}} . This is the total force frictional force needed to hold the box in place. To verify the static friction can provide enough force we can continue calculate the MAX static friction force. This number should be greater than 150.8 \, \text{N} .
5 f_{max} = \mu_s F_{\text{normal}} Maximum force of static friction that can be applied.
6 F_{\text{normal}} = F_{\text{gravity}} \cos(\theta) + F_{\text{applied}} The normal force is affected by both the gravitational perpendicular force and the additional applied force of 59 N, which acts perpendicular to the slope, reducing the normal force.
6 F_{\text{normal}} = 245 \cos(38.0^\circ) + 59 = 252.1 \, \text{N} Calculation of the normal force considering the cosine of the incline angle and the applied force.
7 f_s = \mu_s F_{\text{normal}} Calculation of the max static friction force, where \mu_s is the coefficient of static friction between the crate and the pavement.
8 f_s = 0.599 \times 252.1 = 151 \, \text{N} The magnitude of the maximum frictional force is 151 \, \text{N} which is greater than the 150.8 \, \text{N} . to keep it in place. Thus the scenario is possible.

## Need Help? Ask Phy To Explain This Problem

Phy can also check your working. Just snap a picture!

Simple Chat Box

## See how Others Did on this question | Coming Soon

150.8 \, \text{N}} .

## Continue with

By continuing you (1) agree to our Terms of Sale and Terms of Use and (2) consent to sharing your IP and browser information used by this site’s security protocols as outlined in our Privacy Policy.

Enjoying UBQ? Share the 🔗 with friends!

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!

## Phy Pro

The most advanced version of Phy. Currently 50% off, for early supporters.

## \$11.99

per month

Billed Monthly. Cancel Anytime.

Trial  –>  Phy Pro