Question

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? ...?

Answer 1

If there isn't any force then the normal contact force will be 


N=m*g=7.5*9.81=73.58N 

which is 73.58-23=50.58N less 

so, there the person must pull at 23 degree upward 

break down the tension in two components, vertical and horizontal. 


vertical tension= 50.58=T*sin23 

T=50.58/sin23=129.45N

Answer 2

Answer:

Total tension in the person's arm = 58.82 N

Explanation:

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