According to Newton's second law of motion, resultant force on a particle is in the direction of acceleration of the particle.
Two forces F₁ and F₂ are acting on a particle as shown in the figure. The acceleration of the particle is along X-axis. The value of F₁ (in newton) is

Question

According to Newton's second law of motion, resultant force on a particle is in the direction of acceleration of the particle.

Two forces F₁ and F₂ are acting on a particle as shown in the figure. The acceleration of the particle is along X-axis. The value of F₁ (in newton) is

Infinity Learn · Accepted Answer

Since, acceleration of particle is along X-axis. So, net forces will be along X-axis.

$Hence, \sum F_{Y} = 0$

$∴ F_{2} \sin 37^{\circ} - F_{1} \sin 30^{\circ} = 0 or \frac{3}{5} F_{2} = \frac{F_{1}}{2}$

$∴ F_{1} = \frac{6}{5} \times 5 = 6 N$

Newton's laws of motion

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According to Newton's second law of motion, resultant force on a particle is in the direction of acceleration of the particle.
Two forces F₁ and F₂ are acting on a particle as shown in the figure. The acceleration of the particle is along X-axis. The value of F₁ (in newton) is

Since, acceleration of particle is along X-axis. So, net forces will be along X-axis.
$Hence, \sum F_{Y} = 0$
$∴ F_{2} \sin 37^{\circ} - F_{1} \sin 30^{\circ} = 0 or \frac{3}{5} F_{2} = \frac{F_{1}}{2}$
$∴ F_{1} = \frac{6}{5} \times 5 = 6 N$

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Newton's laws of motion

Remember concepts with our Masterclasses.

According to Newton's second law of motion, resultant force on a particle is in the direction of acceleration of the particle.Two forces F1 and F2 are acting on a particle as shown in the figure. The acceleration of the particle is along X-axis. The value of F1 (in newton) is

Since, acceleration of particle is along X-axis. So, net forces will be along X-axis. Hence, ∑FY=0∴F2sin⁡37∘−F1sin⁡30∘=0 or 35F2=F12∴ F1=65×5=6N

Ready to Test Your Skills?

free study material

According to Newton's second law of motion, resultant force on a particle is in the direction of acceleration of the particle.
Two forces F₁ and F₂ are acting on a particle as shown in the figure. The acceleration of the particle is along X-axis. The value of F₁ (in newton) is

Since, acceleration of particle is along X-axis. So, net forces will be along X-axis.
$Hence, \sum F_{Y} = 0$
$∴ F_{2} \sin 37^{\circ} - F_{1} \sin 30^{\circ} = 0 or \frac{3}{5} F_{2} = \frac{F_{1}}{2}$
$∴ F_{1} = \frac{6}{5} \times 5 = 6 N$