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A block with a mass of 6.0 kg is
held in equilibrium on an incline
of angle θ = 30.0° by a horizontal
force, F, as shown in the figure.
Find the magnitudes of the
normal force on the block and of F. (Ignore friction.)

Respuesta :

onyebuchinnaji

(a) The normal force on the block is 50.92 N.

(b) The horizontal force on block keeping it in equilibrium is 29.4 N.

The given parameters;

  • mass of the block, m = 6 kg
  • angle of inclination, θ = 30.0°

The normal force on the block is calculated as follows;

[tex]F_n = W \times cos(\theta)[/tex]

where;

  • W is the weight of the block

[tex]F_n = mg \times cos(\theta)\\\\F_n = 6 \times 9.8 \times cos(30)\\\\F_n = 50 .92 \ N[/tex]

The horizontal force on block keeping it in equilibrium is calculated as follows;

[tex]F- F_x = 0\\\\F-mgsin\theta= 0\\\\F = mgsin\theta\\\\F = 6 \times 9.8 \times sin(30)\\\\F = 29.4 \ N[/tex]

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