Physics, 26.09.2019 07:50 daliakreynin

At the end of a factory production line, boxes start from rest and slide down a 30 ∘ ramp 5.8 m long. part a if the slide is to take no more than 3.9 s , what's the maximum allowed frictional coefficient

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The component of weight(mg) that is responsible for the motion of boxes on ramp is:
$m*g*sin \alpha$

Where m = mass of the boxes.
g = Acceleration due to gravity = 9.8 $m/s ^{2}$
$\alpha$ = The angle the ramp makes with the ground. In this case it is 30°.

Since the frictional force is:
$F_{f} =$ μ*N.

Where,
μ = Frictional Coefficient
N = Normal to the ramp = $m*g*cos\alpha$

Therefore, the frictional force becomes = $F_{f}$ = μ* $m*g* cos \alpha$

Apply Newton's second law we would get:
$m*g*sin \alpha$ - μ* $m*g* cos \alpha$ = m*a

= $g*(sin \alpha -$ μ $cos \alpha )$ -- (A)

Now according to equation of motion:
$x = x_{o} + v_{o}*t + 1/2 * a * t^{2}$

Where x = 5.8m
$x_{o}$ = 0
$v_{o}$ = 0
$t^{2}$ = 10.24

Plug in the value in the above equation you would get:

= 1.1328 $m/s ^{2}$

Plug in

in equation (A) and solve for μ, you would get,
μ = 0.4439

Answer from: DaisyHargrove23

We can solve the problem by analyzing the forces acting along the direction of the inclined plane. There are two forces: the first one is the component of the weight parallel to the inclined plane, which is
$mg \sin \alpha$
where g=9.81 m/s^2

and $\alpha=30^{\circ}$ . This force points along the direction of the motion (down). The second one is the frictional force, acting against the motion, as well along the inclined plane:
$-m g \mu_D$
where $\mu_D$ is the dynamic frictional coefficient, and we have to find the value of this. The negative sign means the force is acting against the direction of the motion.

Newton's second law states that the resultant of the forces acting on the object is equal to the product between the mass m and the acceleration a:
$mg \sin \alpha - mg \mu_D = ma$
So we can write the acceleration as
$a=g \sin \alpha - g \mu_D$

Now, we can use the law of motion along the direction of the inclined plane. For an uniformly accelerated motion, we have
$S= \frac{1}{2} a t^2$
Assuming the object is initially at rest. But S, in our problem, is exactly the lenght of the plane:
S=L=5.8 m

So we can substitute a inside the formula and we get
$L= \frac{1}{2} (g \sin \alpha - g \mu_D) t^2$
From which we get
$\mu_D = \sin \alpha - \frac{2L}{gt^2}$
and using the time given by the problem, t=3.9 s

and

, we find
$\mu_D = 0.422$

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