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You push on a cart (18.0kg) at a 30 degree below horizontal angle. The coefficient of kinetic friction between the chair and the floor is 0.625. What is the value of the angled force that will keep the chair moving at a constant velocity?

Respuesta :

Given that force is applied at an angle of 30 degree below the horizontal

So let say force applied if F

now its two components are given as

[tex]F_x = Fcos30[/tex]

[tex]F_y = Fsin30[/tex]

Now the normal force on the block is given as

[tex]N = Fsin30 + mg[/tex]

[tex]N = 0.5F + (18\times 9.8)[/tex]

[tex]N = 0.5F + 176.4[/tex]

now the friction force on the cart is given as

[tex]F_f = \mu N[/tex]

[tex]F_f = 0.625(0.5F + 176.4)[/tex]

[tex]F_f = 110.25 + 0.3125F[/tex]

now if cart moves with constant speed then net force on cart must be zero

so now we have

[tex]F_f + F_x = 0[/tex]

[tex]Fcos30 - (110.25 + 0.3125F) = 0[/tex]

[tex]0.866F - 0.3125F = 110.25[/tex]

[tex]F = 199.2 N[/tex]

so the force must be 199.2 N

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