To solve the problem we will draw the free body diagram first and then calculate the normal force.
The force applies to the crate is 49.05 Newton.
Given to us
- Mass of the crate, m = 50 kg
- coefficient of kinetic friction between the crate and the floor, μ = 0.1.
Free Body Diagram
Let's draw the Free Body Diagram of the crate first, as shown below, in the image.
Now,
Sum of all vertical force = 0,
\(\sum F_y = 0\)
\(\text{Weight of the crate} = \text{Normal Force}\)
\(m\times g = N\\N = m\times g\\N = 50 \times 9.81\\\)
Sum of all the Horizontal force = 0,
\(\sum F_x =0\\\)
\(\rm{Force = Frictional\ Force\)
\(F = \mu \times N\\F = 0.1 \times 50 \times 9.81\\F = 49.05\)
Hence, the force applies to the crate is 49.05 Newton.
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