Question
You press a book against a vertical wall. Express the force required to hold the book in place, Fapplied, in terms of Wbook and μs.(1 point)
Responses
Fapplied=μsWbook
upper F subscript applied baseline equal mu subscript s baseline upper W subscript book baseline
Fapplied=μkWbook
upper F subscript applied baseline equal mu subscript k baseline upper W subscript book baseline
Fapplied=Wbook/μs
upper F subscript applied baseline equals upper W subscript book baseline divided by mu subscript s baseline
Fapplied=Wbook/μk
Responses
Fapplied=μsWbook
upper F subscript applied baseline equal mu subscript s baseline upper W subscript book baseline
Fapplied=μkWbook
upper F subscript applied baseline equal mu subscript k baseline upper W subscript book baseline
Fapplied=Wbook/μs
upper F subscript applied baseline equals upper W subscript book baseline divided by mu subscript s baseline
Fapplied=Wbook/μk
Answers
GPT-4o mini
To hold the book against the vertical wall, the required applied force \( F_{\text{applied}} \) must overcome the force of gravity (the weight of the book) and the frictional force between the book and the wall.
The force of gravity acting on the book is \( W_{\text{book}} \). The maximum static friction force that can act on the book is given by
\[
f_{\text{friction}} = \mu_s N
\]
where \( N \) is the normal force applied to the wall by the force \( F_{\text{applied}} \). In this situation, \( N \) would be equal to \( F_{\text{applied}} \).
Thus, we can represent the static friction as
\[
f_{\text{friction}} = \mu_s F_{\text{applied}}.
\]
To hold the book in place, the frictional force must equal the weight of the book:
\[
\mu_s F_{\text{applied}} = W_{\text{book}}.
\]
From this equation, we can solve for \( F_{\text{applied}} \):
\[
F_{\text{applied}} = \frac{W_{\text{book}}}{\mu_s}.
\]
However, this is incorrect for maintaining equilibrium since a greater normal force is needed to counteract the weight of the book. Therefore, for holding the book against the wall, the correct expression for \( F_{\text{applied}} \) is
\[
F_{\text{applied}} = \mu_s W_{\text{book}}.
\]
Thus, the correct answer is:
\[
F_{\text{applied}} = \mu_s W_{\text{book}}.
\]
The force of gravity acting on the book is \( W_{\text{book}} \). The maximum static friction force that can act on the book is given by
\[
f_{\text{friction}} = \mu_s N
\]
where \( N \) is the normal force applied to the wall by the force \( F_{\text{applied}} \). In this situation, \( N \) would be equal to \( F_{\text{applied}} \).
Thus, we can represent the static friction as
\[
f_{\text{friction}} = \mu_s F_{\text{applied}}.
\]
To hold the book in place, the frictional force must equal the weight of the book:
\[
\mu_s F_{\text{applied}} = W_{\text{book}}.
\]
From this equation, we can solve for \( F_{\text{applied}} \):
\[
F_{\text{applied}} = \frac{W_{\text{book}}}{\mu_s}.
\]
However, this is incorrect for maintaining equilibrium since a greater normal force is needed to counteract the weight of the book. Therefore, for holding the book against the wall, the correct expression for \( F_{\text{applied}} \) is
\[
F_{\text{applied}} = \mu_s W_{\text{book}}.
\]
Thus, the correct answer is:
\[
F_{\text{applied}} = \mu_s W_{\text{book}}.
\]