Question

A 2.95 kg object is subjected to three forces that give it an acceleration a with arrow = −(8.00 m/s2)i hat + (6.00 m/s2)j. If two of the three forces are F with arrow1 = (36.6 N)i hat + (16.2 N)j and F with arrow2 = −(12.0 N)i hat + (8.00 N)j, find the third force.

Answer 1

Answer:$\vec{F_3}=48.19\hat{i}-6.49\hat{j}$

Explanation:

Given

Force $\vec{F_1}=36.6\hat{i}+16.2\hat{j}$

$\vec{F_2}=-12\hat{i}+8\hat{j}$

let us suppose $\vec{F_3}=F\cos \theta \hat{i}+\F\sin \theta \hat{j}$

acceleration

$\vec{a}=-8\hat{i}+6\hat{j}$

Force in x direction

$F_x_{net}=36.6+-12+F\sin \theta$

$a_{x}=\frac{F_x_{net}}{m}=\frac{36.6+-12+F\cos \theta }{2.95}$

$-8\times 2.95=36.6-12+F\cos \theta$

$-23.6=24.6+F\cos \theta$

$F\cos \theta =-48.2$----1

$a_{y}=\frac{F_y_{net}}{m}=\frac{16.2+8+F\sin \theta }{2.95}$

$2.95\times 6=16.2+8+F\sin \theta$

$17.7=24.2+F\sin \theta$

$F\sin \theta =-6.5$-----2

squaring and adding 1 & 2

$F^2=2365.49$

F=48.63 N

substitute F in 2

$sin\theta =-0.1336$

$\theta =-7.677$

thus $\vec{F_3}=48.63\cos (-7.677)\hat{i}+48.63\sin (-7.677)\hat{j}$

$\vec{F_3}=48.19\hat{i}-6.49\hat{j}$