Home Physics Consider the static equilibrium diagram here. What is the angle F1 must make with the horizontal? Consider the static equilibrium diagram here. What is the angle F1 must make with the horizontal?Physics Karateprincess1October 23, 20212 CommentsConsider the static equilibrium diagram here. What is the angle F1 must make with the horizontal?
ANSWER[tex]\theta=35\degree[/tex]EXPLANATIONSince the body is in equilibrium, total upward forces must equal total downward force.Also the net horizontal forces acting on the body must be zero.We need to resolve [tex]F_1[/tex] into vertical and horizontal components.The horizontal component is[tex]x=F_1\cos\theta[/tex].The vertical component is[tex]y=F_1\sin\theta[/tex].Equating the up force to the downward forces gives,[tex]F_1\sin\theta + 20N=60N[/tex].This implies that,[tex]F_1\sin\theta =60N-20N[/tex].[tex]F_1\sin\theta=40N...eqn1[/tex]Also the horizontal forces must be equal.[tex]F_1\cos\theta=57N...eqn2[/tex].Dividing equation (1) by equation (2) gives,[tex]\frac{F_1\sin\theta}{F_1\cos\theta}=\frac{40}{57}[/tex].[tex]\Rightarrow \tan\theta=0.70175[/tex][tex]\Rightarrow \theta=tan^{-1}(0.70175)[/tex][tex]\Rightarrow \theta=35.0594[/tex].Therefore the given angle that [tex]F_1[/tex] must make with the horizontal is approximately 35° to the nearest degree.Reply
ANSWER
[tex]\theta=35\degree[/tex]
EXPLANATION
Since the body is in equilibrium, total upward forces must equal total downward force.
Also the net horizontal forces acting on the body must be zero.
We need to resolve [tex]F_1[/tex] into vertical and horizontal components.
The horizontal component is
[tex]x=F_1\cos\theta[/tex].
The vertical component is
[tex]y=F_1\sin\theta[/tex].
Equating the up force to the downward forces gives,
[tex]F_1\sin\theta + 20N=60N[/tex].
This implies that,
[tex]F_1\sin\theta =60N-20N[/tex].
[tex]F_1\sin\theta=40N...eqn1[/tex]
Also the horizontal forces must be equal.
[tex]F_1\cos\theta=57N...eqn2[/tex].
Dividing equation (1) by equation (2) gives,
[tex]\frac{F_1\sin\theta}{F_1\cos\theta}=\frac{40}{57}[/tex].
[tex]\Rightarrow \tan\theta=0.70175[/tex]
[tex]\Rightarrow \theta=tan^{-1}(0.70175)[/tex]
[tex]\Rightarrow \theta=35.0594[/tex].
Therefore the given angle that [tex]F_1[/tex] must make with the horizontal is approximately 35° to the nearest degree.
A. 35 on plato, got 100 on the test 🙂