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At a construction site, a 68.0 kg bucket of concrete hangs from a light (but strong) cable that passes over a light friction-free pulley and is connected to an 85.0 kg box on a horizontal roof (see the figure (Figure 1)). The cable pulls horizontally on the box, and a 46.0 kg bag of gravel rests on top of the box. The coefficients of friction between the box and roof are shown. The system is not moving.

Respuesta :

Answer:

Gravel Box Ms = 0.700 MK = 0.400 Concrete

Explanation:

Following are the calculation to the friction force:

  • Please see the diagram below for an illustration of the forces acting on objects.
  • Therefore, m denotes the mass of a gravel bag, and M denotes the overall mass of the box.
  • Whenever the system is stationary, the friction force on the box equals the tension in the string.

The substring tension force is provided as follows:  

[tex]\to T-m_{1}g=0 \\\\\to T = m_{1}g \\\\[/tex]

       [tex]= (68\ kg) (9.81 \ \frac{m}{s^2} ) (\frac{ 1\ N}{ 1\ kg \cdot \frac{m}{s^2}}) \\\\ = (68 ) (9.81 ) ( 1\ N) \\\\= 667.08\ N \\\\[/tex]

Therefore, the friction force on the box is "667 N".

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