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A child pulls a rope attached to a stone with a force of $60 \mathrm{~N}$. The rope makes an angle of 60 degrees to the ground. What is the effective force along the horizontal direction?
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Verified Answer
The correct answer is:
$30 \mathrm{~N}$
CALCULATION:
The horizontal component is responsible to move the stone along the ground.
$\vec{F}=F_x=F \cos \theta$
$F_X=60 \cos 60^{\circ}=30 \mathrm{~N}$
Hence, the effective value of the pull tending to move the stone on the ground is $30 \mathrm{~N}$.
CONCEPT:
Resolution of vectors into components: We have a vector $(F)$ where the magnitude of the vector is $\mathrm{F}$ and the angle with horizontal is $\theta$.
The vector has two components: 1. Vertical component and 2. Horizontal component
Vertical component $\left(F_y\right)=F \sin \theta$
Horizontal component $\left(\mathrm{F}_{\mathrm{x}}\right)=\mathrm{F} \operatorname{Cos} \theta$
The horizontal component is responsible to move the stone along the ground.
$\vec{F}=F_x=F \cos \theta$
$F_X=60 \cos 60^{\circ}=30 \mathrm{~N}$
Hence, the effective value of the pull tending to move the stone on the ground is $30 \mathrm{~N}$.
CONCEPT:
Resolution of vectors into components: We have a vector $(F)$ where the magnitude of the vector is $\mathrm{F}$ and the angle with horizontal is $\theta$.
The vector has two components: 1. Vertical component and 2. Horizontal component
Vertical component $\left(F_y\right)=F \sin \theta$
Horizontal component $\left(\mathrm{F}_{\mathrm{x}}\right)=\mathrm{F} \operatorname{Cos} \theta$
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