Question

A child pulls a box with a force $200\;N$ at an angle of $60^{\circ}$ above the horizontal. Then the horizontal and vertical components of the force are:

& A.100N,175N \\\ & B.86.6N,100N \\\ & C.100N,86.6N \\\ & D.100N,0N \\\ \end{aligned}$$

Answer 1

Here we have a force which is inclined at some angle. Then we can resolve the force into the vertical component and the horizontal component. Also, the horizontal component $Fcos\theta=F_{X}$ and the vertical component $Fsin\theta=F_{Y}$

Formula used: $Fcos\theta=F_{X}$ and $Fsin\theta=F_{Y}$

Complete step by step answer:
We know that force is defined as a pull or a push of any given object. Here it is given that the force exerted to pull the box is $F=200N$ and makes an angle $60^{\circ}$as shown in the diagram.

Since force is a vector we know that it has both direction and magnitude. Here since it makes some angle with the vertical it can be resolved into two perpendicular components, namely the vertical component and the horizontal component. From basic trigonometry, we can find the magnitude of the vector or the angle it makes. Using the phasor diagram we can find the direction or the orientation of the vector.
Then the $F_{X}$ is the horizontal component is given as $Fcos\theta=F_{X}$. While the $F_{Y}$ is the vertical component is given as $Fsin\theta=F_{Y}$

Here , $F=200N$ and makes an angle $\theta=60^{\circ}$ then we get,$F_{X}$ is the horizontal component, then $F_{X}=Fcos60=200\times\dfrac{1}{2}=100N$
Similarly, $F_{Y}$ is the vertical component, then $F_{Y}=Fsin60=200\times\dfrac{\sqrt3}{2}=100\sqrt3N=173.2N$ which is approximately 175N.

So, the correct answer is “Option A”.

Note: Note that the options C and B are similar. Since the question asks the horizontal and vertical components of the force hence option C is correct and not B. Thus it is suggested to read the question well and understand it before answering. This is a very simple question and failing to read the question will lead to the wrong answer.