- #1

Kalix

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## Homework Statement

Question: If an 80kg object experiences a normal force of 510N, and an unknown angled pull force. The coefficient of friction is 0.33. If the object moves with a constant velocity, what is the direction (angle) and magnitude (size) of the pull force?

m=80kg

a=0m/s

Fx=FcosΘ

Fy=FsinΘ

μ(mu)=0.33

Fk=510N x 0.33

g=-9.81

weight=80kg x 9.81

## Homework Equations

Fx-fk=max (the "x" is a coefficient)

Fy-fk=may (the "y" is a coefficient)

Fs=μs x Fn

Fk=μk x Fn

## The Attempt at a Solution

First I set up a simultaneous equation using Fx-fk=max and Fy-fk=may. This is what I got.

80kg+0m/s=FcosΘ -(510N)(0.33)

80kg+0m/s=FsinΘ -(80kg)(9.81)

This is where I get stuck. I am really bad at solving simultaneous equations.