Question

A 15.5 kg block is pulled by two forces. The first is 11.8 N at a 53.7 angle and the second is 22.9 at a -15.8 angle. What is the magnitude and direction of the acceleration?

Answer 1

Answer:

$1.88 m/s^2$ at $6.5^{\circ}$

Explanation:

We need to calculate the components of the resultant force on both the x (horizontal) and y (vertical) direction.

Components of the first force F1:

$F_{1x} =(11.8) cos (53.7^{\circ})=7.0 N\\F_{1y} = (11.8) sin (53.7^{\circ})=9.5 N$

Components of the second force F2:

$F_{2x} =(22.9) cos (-15.8^{\circ})=22.0 N\\F_{2y} = (22.9) sin (-15.8^{\circ})=-6.2 N$

So the components of the resultant force are

$R_x = F_{1x}+F_{2x}=7.0+22.0 = 29.0 N\\R_y = F_{1y}+F_{2y} = 9.5+(-6.2)=3.3 N$

So the magnitude of the resultant force is

$F=\sqrt{(29.0)^2+(3.3)^2}=29.2 N$

And the direction is

$\theta = tan^{-1} (\frac{R_y}{R_x})=tan^{-1} (\frac{3.3}{29.0})=6.5^{\circ}$

The magnitude of the acceleration can be found by using Newton's second law:

$a=\frac{F}{m}=\frac{29.2 N}{15.5 kg}=1.88 m/s^2$

while the direction is the same as the resultant force, $6.5^{\circ}$.