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**1. Homework Statement**

A block of weight 20 N (m = 2 kg) sits on an plane inclined at 37°.

*g*= 10 m/s

^{2}(for simplicity).

**a)**Calculate the value of the weight components.

**b)**Calculate the acceleration assuming no friction.

**c)**Calculate the acceleration assuming μ

_{k}= 0.125.

**d)**What value of μ

_{k}is needed to make the block slide down at constant velocity?

**e)**How great must μ

_{s}be to prevent sliding?

**f)**In what way to the answers to d) and e) depend on the block's mass?

**2. Homework Equations**

Taking the incline as the x-axis,

Fw

_{x}= Fwsinθ

Fw

_{y}= Fwcosθ

∑F

_{x}: Fwsin37° - F

_{F}= ma

_{x}

∑F

_{y}: Fwcos37° + F

_{N}= ma

_{y}

tan θ = tan 37° = 0.75

**3. The Attempt at a Solution**

a)Fw

a)

_{x}= 20sin37° =

__12 N__

Fw

_{y}= 20cos37° =

__16 N = F__

_{N}**b)**∑F

_{x}: 20sin37° = 2a, ∴

__a = 6 m/s__

^{2}**c)**∑F

_{x}: 20sin37° - F

_{F}= 2a,

12 - ((μ

_{k})(F

_{N})) = 2a

12 - ((0.125)(15.97) = 2a

12 - 2 = 2a

__a = 10/2 = 5 m/s__

^{2}**d)**Constant velocity implies a = 0.

∑F

_{x}: Fwsin37° - F

_{F}= 0, ∴F

_{F}= 12 N -- The force of friction balances the component of gravity pulling the box down the incline.

If F

_{F}= 12 N,

__μ__

_{k}= F_{F}/F_{N}= 12/16 = 0.75**e)**I know that the coefficient of kinetic friction (μ

_{k}) is usually lower than the coefficient of static friction (μ

_{s}). If I were told what force was required to disrupt equilibrium (no motion), μ

_{s}would be given by (Force required/F

_{N}). Without knowing this force I'm not sure how to proceed next.

**f)**My suspicion is that these values do not depend on mass at all, since acceleration is zero in each of the cases described in d) and e) (constant velocity and no motion until the friction force resisting motion is overcome.

Any and all help is greatly appreciated.