An object with a mass of $7 k g$ $7 kg$ is on a ramp at an incline of $\frac{π}{12}$ $pi/12$ . If the object is being pushed up the ramp with a force of $8 N$ $8 N$ , what is the minimum coefficient of static friction needed for the object to remain put?

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Answer 1 · 2017-06-07T18:27:37

THe coefficient of static friction is $= - 0.15$ $=-0.15$

Taking the direction up and parallel to the plane as positive $↗^{+}$ $↗^+$

The coefficient of static friction is $μ_{s} = \frac{F_{r}}{N}$ $mu_s=F_r/N$

The normal force is $= N = m g cos θ$ $=N=mgcostheta$

Then the net force on the object is

$F = F_{r} + W sin θ$ $F=F_r+Wsintheta$

$= F_{r} + m g sin θ$ $=F_r+mgsintheta$

$= μ_{s} N + m g sin θ$ $=mu_sN+mgsintheta$

$= m μ_{s} g cos θ + m g sin θ$ $=mmu_sgcostheta+mgsintheta$

$= m g (μ_{s} cos θ + sin θ)$ $=mg(mu_scostheta+sintheta)$

So,

$\frac{F}{m g} = (μ_{s} cos θ + sin θ)$ $F/(mg)=(mu_scostheta+sintheta)$

$μ_{s} cos θ = \frac{F}{m g} - sin θ$ $mu_scostheta=F/(mg)-sintheta$

$μ_{s} = \frac{1}{cos θ} \cdot (\frac{F}{m g} - sin θ)$ $mu_s=1/costheta*(F/(mg)-sintheta)$

$= \frac{1}{cos (\frac{π}{12})} \cdot (\frac{8}{7 \cdot 9.8} - sin (\frac{π}{12}))$ $=1/cos(pi/12)*(8/(7*9.8)-sin(pi/12))$

$= - 0.15$ $=-0.15$

An object with a mass of 7kg7 kg is on a ramp at an incline of π12pi/12 . If the object is being pushed up the ramp with a force of 8N 8 N, what is the minimum coefficient of static friction needed for the object to remain put?