Question

Two ropes are attached to a 40-kg object. The first rope applies a force of 50 N and the second, 40 N. If the two ropes are perpendicular to each other, what is the resultant magnitude of the acceleration of the object?

Answer 1

To solve this problem we will use the concepts related to the resulting Vector Force product of two components, that is,

$|\vec{F}| = \sqrt{F_x^2+F_y^2}$

If we take the Force of 50 N as the force in the X direction and the Force of 40 N in the Y direction we will have to:

$|\vec{F}| = \sqrt{F_x^2+F_y^2}$

$|\vec{F}| = \sqrt{(50)^2+(40)^2}$

$|\vec{F}| = 64.03N$

Finally, since Newton's second law, acceleration can be determined as

$F = ma$

$a = \frac{F}{m}$

$a = \frac{ 64.03}{40}$

$a = 1.6m/s^2$

Therefore the resultant magnitude of the acceleration of the object is $1.6m/s^2$