Force question! (physics)

You pull a heavy 300 kg crate across a level floor with a rope that makes a 20 degree angle with the floor. The crate moves horizontally with constant velocity. The coefficient of kinetic friction is 0.200. What is the magnitude of your force required to pull it?

Horizontal forces give us the equation:

ΣF(x) = 0 = Tcosθ - µn------------------>(1)

The normal force is found from the vertical equations:

ΣF(y) = 0 = n - mg + Tsinθ
n = mg - Tsinθ------------------------->(2)

Subbing (2) into (1) and solving for T:

0 = Tcosθ - µ(mg - Tsinθ)
T = µmg / (cosθ + µsinθ)
= 0.200(300kg)(9.81m/s²) / (cos20.0° + 0.200sin20.0°)
= 584N (rounded)

Hope this helps.

The vertical force balance is that between gravity on the one hand, and vertical component of pull (Fp) and normal force (Fn) on the other hand:

Fn + Fp sin(20) = m g

The horizontal force balance (uniform motion, so net force equals zero) is between horizontal component of pull and kinetic friction (μ Fn):

Fp cos(20) = μ Fn

These are two equations for the two unknowns Fp and Fn.

From the second equation we know:

Fn = Fp cos(20) / μ

Substituting that in the first equation gives:

Fp cos(20) / μ + Fp sin(20) = m g

Fp ( cos(20) / μ + sin(20) )= m g

Fp = m g / ( cos(20) / μ + sin(20) )


Thus:

Fp = 300 kg * 9.81 m/s^2 / ( cos(20) / 0.200 + sin(20) )
= 584 N