A computer base unit of mass 7.5 kg is dragged along a smooth desk. If the normal contact force is 23N and the tension in the arm of the person dragging it acts at 23 degrees to the horizontal, then what is the total tension in the person's arms? ...?

Normal Contact Force: Normal contact force is the vertical component of the contact force that is perpendicular to the surface that an object makes contact with.

As shown in fig.1 in the diagram on the attached file below.

The Normal contact force (R) = Tsinθ = The vertical component of T

Where T = Total tension in the person's arm (N), θ = angle of inclination to the horizontal.

∴ R = Tsinθ

T = R/sinθ where R= 23N, θ =23°

T = 23/sin23° = 23/0.391

T = 58.82 N

∴ Total tension in the person's arm = 58.52 N

okpalawalter8 okpalawalter8 · Answer 2 · 2021-10-29T15:13:01+02:00

We have that from the Question, it can be said that the total tension in the person's arms is

Ft=129N

From the Question we are told

A computer base unit of mass 7.5 kg is dragged along a smooth desk. If the normal contact force is 23N and the tension in the arm of the person dragging it acts at 23 degrees to the horizontal, then what is the total tension in the person's arms? ...?

Generally the equation for Horizontal motion is mathematically given as

[tex]\sumFx=FtCos \theta\\\\\sumFx=ma[/tex]

Generally the equation for vertical motion is mathematically given as

[tex]Ftsin\theta+fn=mh\\\\Therefore\\\\Ft=\frac{7.5*9.8-23}{sin23}[/tex]

Ft=129N

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