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Two dimensional dynamics often involves solving for two unknown quantities in two separate equations describing the total force. The block in (Figure 1) has a mass m=10kg
and is being pulled by a force F
on a table with coefficient of static friction μs=0.3
. Four forces act on it:

The applied force F
(directed θ=30∘
above the horizontal).
The force of gravity Fg=mg
(directly down, where g=9.8m/s2
).
The normal force N
(directly up).
The force of static friction fs
(directly left, opposing any potential motion).
If we want to find the size of the force necessary to just barely overcome static friction (in which case fs=μsN
), we use the condition that the sum of the forces in both directions must be 0. Using some basic trigonometry, we can write this condition out for the forces in both the horizontal and vertical directions, respectively, as:

Fcosθ−μsN=0

Fsinθ+N−mg=0

In order to find the magnitude of force F
, we have to solve a system of two equations with both F
and the normal force N
unknown. Use the methods we have learned to find an expression for F
in terms of m
, g
, θ
, and μs
(no N
).

Express your answer in terms of m
, g
, θ
, and μs
.

Respuesta :

To find the magnitude of force \( F \) in terms of \( m \), \( g \), \( \theta \), and \( \mu_s \), we'll solve the system of equations representing the sum of forces in the horizontal and vertical directions:

\[ F \cos \theta - \mu_s N = 0 \]
\[ F \sin \theta + N - mg = 0 \]

First, we'll solve the first equation for \( N \):

\[ N = \frac{F \cos \theta}{\mu_s} \]

Then, we'll substitute this expression for \( N \) into the second equation:

\[ F \sin \theta + \frac{F \cos \theta}{\mu_s} - mg = 0 \]

Now, we'll solve for \( F \):

\[ F \sin \theta + \frac{F \cos \theta}{\mu_s} = mg \]
\[ F \left( \sin \theta + \frac{\cos \theta}{\mu_s} \right) = mg \]
\[ F = \frac{mg}{\sin \theta + \frac{\cos \theta}{\mu_s}} \]

Therefore, the expression for \( F \) in terms of \( m \), \( g \), \( \theta \), and \( \mu_s \) is:

\[ F = \frac{mg}{\sin \theta + \frac{\cos \theta}{\mu_s}} \]

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