# Magnitude of force acting on an object

#### hunter457

1. The problem statement, all variables and given/known data

A 1.84kg mass is observed to accelerate at 10.66m/s^2 in a direction 28.8 degrees north of east. There is a force directed north acting on the object and a force directed east acting on the object. What is the magnitude of the force acting east on the object?

2. Relevant equations

F=ma, Fx=ma, Fy=ma

3. The attempt at a solution

*FA=force directed north, FB=force directed east, FN=normal force

Fx=ma
-mgsinθ+FAcosα+FBsinβ=ma
-(1.84)(9.8)(sin28.8)+FA(cos61.2)+FB(sin61.2)=(1.84)(10.66)
-8.68698+0.48175FA+0.8763FB=19.6144
0.48175FA+0.8763FB=28.30138
FA=(28.30138-0.8763FB)/0.48175

Fy=ma
FN+FAsinα-FBcosβ-mgcosθ=ma
FN+FAsin61.2-FBcos61.2-(1.84)(9.8)(cos28.8)=(1.84)(10.66)
FN+0.8763FA-0.48175FB-15.8=19.6144
FN+0.8763FA-0.48175FB=35.416
FA=(35.416-FN+0.48175FB)/0.8763

If I put the two equations together to cancel out FA, I still have two unknowns. I don't know what I'm doing wrong. Maybe my FBD was wrong?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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#### Pengwuino

Gold Member
Re: Forces/Acceleration

Does this problem explicitly say gravity is even involved? This sounds like a simple vector problem.

#### hunter457

Re: Forces/Acceleration

You're right. I didn't need to use forces to solve this:

F = m*a
= 1.84*10.66
= 19.6 N

If I draw a line from the origin 28.8deg north of east and derive the components, then Fx=the force eastward:

Fx = 19.6 *cos(28.8) = 17.2 N

#### Redbelly98

Staff Emeritus
Homework Helper
Re: Forces/Acceleration

Looks good

p.s. Welcome to Physics Forums.

#### hunter457

Re: Forces/Acceleration

Thank you. I know that I'll be using this forum a lot more

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