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A push of magnitude p acts on a box of weight w as shown in the figure. the push is directed at an angle θ below the horizontal, and the box remains a rest. the box rests on a horizontal surface that has some friction with the box. the friction force on the box due to the floor is equal to

Respuesta :

AbhiGhost
The friction force is equal to the horizontal component of applied force 'p'. This horizontal component of force is pcosθ( θ = angle made by force with horizontal ). 
Hence the frictional force is equal to pcosθ.

Answer:

[tex]F_f = P cos\theta[/tex]

Explanation:

Since applied force is at an angle with the horizontal so we can find the two components of the force on it

[tex]F_x = P cos\theta[/tex]

[tex]F_y = P sin\theta[/tex]

now the normal force due to floor is given as

[tex]F_n = P sin\theta + W[/tex]

Also we know that the box is not moving but it remains at rest

so here we can say that friction force due to floor will counterbalance the horizontal applied force on the box

[tex]F_x = F_f[/tex]

[tex]F_f = P cos\theta[/tex]

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