Question

A force \(\vec{F_1}\) of magnitude \(4\;N\) acts on an object of mass \(1\;kg\), at origin in a direction \(30^\circ\) above the positive \(x\)-axis. A second force \(\vec{F_2}\) of magnitude \(4\;N\) acts on the same object in the direction of the positive \(y\)-axis. The magnitude of the acceleration of the object is nearly

• A. \(6.9\;\text{m s}^{-2}\)
• B. \(7.6\;\text{m s}^{-2}\)
• C. \(4.3\;\text{m s}^{-2}\)
• D. \(8.0\;\text{m s}^{-2}\)

Hint:

When forces act at angles, first resolve them into horizontal and vertical components, then find the resultant using the Pythagoras theorem.

Answer 1

The Correct Option is A

Step 1: Note the two forces acting.
A mass $m = 1\ kg$ feels $\vec{F_1} = 4\ N$ at $30^\circ$ above the $+x$-axis and $\vec{F_2} = 4\ N$ along the $+y$-axis. We want the magnitude of the acceleration, so first we find the net force.
Step 2: Resolve the first force into components.
\[ F_{1x} = 4\cos 30^\circ = 4 \times \tfrac{\sqrt{3}}{2} = 2\sqrt{3}\ N \] \[ F_{1y} = 4\sin 30^\circ = 4 \times \tfrac{1}{2} = 2\ N \]
Step 3: Write the components of the second force.
Since $\vec{F_2}$ points straight up the $y$-axis, $F_{2x} = 0$ and $F_{2y} = 4\ N$.
Step 4: Add the components.
\[ F_x = 2\sqrt{3} + 0 = 2\sqrt{3} \approx 3.46\ N \] \[ F_y = 2 + 4 = 6\ N \]
Step 5: Find the magnitude of the resultant force.
\[ F = \sqrt{F_x^2 + F_y^2} = \sqrt{(3.46)^2 + 6^2} = \sqrt{12 + 36} = \sqrt{48} \approx 6.93\ N \]
Step 6: Apply Newton's second law.
With $m = 1\ kg$, the acceleration is $a = F/m = 6.93/1 \approx 6.9\ \text{m s}^{-2}$, which matches option (1). \[ \boxed{6.9\ \text{m s}^{-2}} \]