Question

A 2 kg object is subjected to three forces that give it an acceleration −→a = −(8.00m/s^2)ˆi + (6.00m/s^2)ˆj. If two of the three forces, are −→F1 = (30.0N)ˆi + (16.0N)ˆj and −→F2 = −(12.0N)ˆi + (8.00N)ˆj, find the third force.

Answer 1

`\vec{F_3}=(-34.0N)\hat{i}+(-12.0N)\hat{j}`

Step-by-step explanation:

You have three forces F1, F2 an F3 that produce the following acceleration:

a = −(8.00m/s^2)ˆi + (6.00m/s^2)ˆj

you know that force F1 and F2 are:

F1 = (30.0N)ˆi + (16.0N)ˆj

F2 = −(12.0N)ˆi + (8.00N)ˆj

and the force F3 is unknown:

F3 = F3x ˆi + F3y ˆj

The second Newton law is given by the following equation:

`\vec{F}=m\vec{a}`

F: the total force = F1 +F2 + F3

m: mass of the object = 2 kg

By the properties of vectors you have:

`\vec{F_1}+\vec{F_2}+\vec{F_3}=m\vec{a}\\\\(30.0-12.0+F_(3x))N\hat{i}+(16.0+8.00+F_(3y))N\hat{j}=(2.0kg)[(-8.00m/s^2)\hat{i}+(6.00m/s^2)\hat{j}]\\\\(18.0+F_(3x))N\hat{i}+(24.0+F_(3y))\hat{j}=(-16.00N)\hat{i}+(12.0N)\hat{j}`

Both x and y component must be equal in the previous equality, then you have:

`18.0N+F_(3x)=-16.00N\\\\F_(3x)=-34.00N\\\\24.0N+F_(3y)=12.0N\\\\F_(3y)=-12.00N`

Hence, the vector F3 is:

`\vec{F_3}=(-34.0N)\hat{i}+(-12.0N)\hat{j}`