Question

Two workers are sliding 440 kg crate across the floor. One worker pushes forward on the crate with a force of 390 N while the other pulls in the same direction with a force of 260 N using a rope connected to the crate. Both forces are horizontal, and the crate slides with a constant speed. What is the crate's coefficient of kinetic friction on the floor

Answer 1

The kinetic coefficient of friction of the crate is 0.1506

Let's assume that both workers are applying positive pressure to the crate. A body cannot change its state of motion whether it is at rest or moves uniformly, according to Newton's First Law (at constant velocity). The magnitude of the friction force must therefore equal the total of the two external forces. The crate's equilibrium equations are as follows:

∑Fₓ = P + T - μ . N = 0

∑$F_{y}$ = N - W = 0

Where:

$P$ - Pushing force, measured in newtons.

$T$ - Tension, measured in newtons.

μ- Coefficient of kinetic friction, dimensionless.

$N$- Normal force, measured in newtons.

W - Weight of the crate, measured in newtons.

The system of equations is now reduced by algebraic means:

P + T - μ . W = 0

We finally clear the coefficient of kinetic friction and apply the definition of weight :

μ $=\frac{P + T}{m.g}$

We know that P = 390 N, T = 260 N, m = 440 kg and $g= 9.807 \frac{m}{s^2}$ ;

μ = 0.1506

Therefore, the kinetic coefficient of friction of the crate is 0.1506

Learn more about kinetic friction here:

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