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## Homework Statement

a 20kg Toaster is on a level surface (Cooefficient of friction is 0.1). A rope applies an unknown force to the toaster at 30° from the horizontal upwards to the right. This force causes the toaster to accelerate at 0.277 m/s^2. What is the magnitude of the applied force.

## Homework Equations

Fnet = m * a

Fn = Fg - Fay

Fax - Ff = Fnet

Ff = μFn (Not sure if thats the symbol for mew)

## The Attempt at a Solution

**Found Fnet**

Fnet = 0.277 * 20

Fnet = 5.54 N

**Friction Along the Floor**

Fn = Fg - Fay

Fay = FaSin30°

Fn = (9.8 * 20) - (FaSin30°)

Fn = 196 - FaSin30°

**Finding Fa**

Fax - Ff = Fnet

Ff = μFn

Fax = FaCos30°

Fax - (0.1*(196 - FaSin30°)) = 5.54 N

Fax - 19.6 - (0.1 * - FaSin30°) = 5.54 N

FaCos30° - 19.6 + (0.1 * FaSin30°) = 5.54 N

FaCos30° + (0.1 * FaSin30°) = 25.11 N

Cos30° + (0.1 * Sin30°) = 25.11 N / Fa

0.866 + (0.1 * 0.5) = 25.11 N / Fa

0.916 = 25.11 / Fa

0.916Fa = 25.11

Fa = 27.4 N

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EDIT: Fax = FaCos30°, making a common variable.

EDIT2: Factored, checking if its right ATM.

EDIT3: YAY :D, Working backwards it works :). Excentric, if anyone can double check my work, would be so awesome.

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