contestada

A 3.06kg wood block is pressed against a vertical wood wall by a force applied at angle ϕ=32.6° to the horizontal. If the block is initially at rest, and μs=0.313, what force would you need to apply to begin moving the box up the wall?

Respuesta :

Let applied force against the wall is F

now the Normal force on the block is given by

[tex]F_n = F cos32.6[/tex]

now friction force on the block will be given as

[tex]F_f = \mu * F_n[/tex]

[tex]F_f = 0.313* (F cos32.6)[/tex]

now net downwards force on the block will be

[tex]F_d= F_f + mg[/tex]

[tex]F_d = 0.313*(Fcos32.6) + 3.06*9.8[/tex]

[tex]F_d = 0.264F + 29.99[/tex]

now this net downward force must be counterbalanced by upwards applied force

[tex]F_u = F_d[/tex]

[tex]Fsin32.6 = 0.264F + 29.99[/tex]

[tex]0.275*F = 29.99[/tex]

[tex]F = 109.1 N[/tex]

so it required 109.1 N force to move it upwards

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