The only given in the dynamics problem is the angle. how can i start with this one?

Question

Piles of snow on slippery roofs can become dangerous projectiles as they melt. Consider a chuck of snow at the ridge of a roof with a pitch of 30o.

a) What is the minimum value of the coefficient of static friction that will keep the snow from sliding down?

B) As the snow begins to melt, the coefficient of static friction decreases and the snow eventually slips. Assuming that the distance from the chuck to the edge of the roof is 5.0m and the coefficient of kinetic friction is 0.20, calculate the speed of the snow chunk when it slides off the roof.

C) If the edge of the roof is 10.0m above ground, what is the speed of the snow when it hits the ground?

Answer 1

<< What is the minimum value of the coefficient of static friction that will keep the snow from sliding down? >>

coefficient of friction = tan A

where

A = 30 degrees

Therefore,

coefficient of friction = tan 30 = 0.5774

<<As the snow begins to melt, the coefficient of static friction decreases and the snow eventually slips. Assuming that the distance from the chuck to the edge of the roof is 5.0m and the coefficient of kinetic friction is 0.20, calculate the speed of the snow chunk when it slides off the roof.>>

acceleration = coeff of kinetic friction * acceleration due to gravity

a = 0.20 * 9.8 = 1.96 m/sec^2

The next formula to use is

Vf^2 - Vo^2 = 2as

where

Vf = velocity of snow as it slides off the roof

Vo = initial velocity = 0 (body is at rest)

a = acceleration (as calculated above) = 1.96 m/sec^2

s = distance that chunk has to travel before falling off the roof = 5 (given)

Substituting appropriate valuse,

Vf^2 - 0 = 2(1.96)(5)



Vf^2 = 19.6

Vf = 4.43 m/sec.



<< If the edge of the roof is 10.0m above ground, what is the speed of the snow when it hits the ground? >>

Use the Law of Conservation of Energy and assuming no wind resistance,

Energy of snow at roof edge = Energy of snow as it hits the round

Energy of snow at roof edge = mgh + (1/2)(m)Vf^2

Energy of snow as it hits the ground = (1/2)mVg^2

where

m = mass of the falling snow

g = as described above = 9.8 m/sec^2

h = distance of snow from ground = 10 m (given)

Vf = velocity of snow as it slides off the roof (as calculated above) = 4.43 m/sec.

Vg = velocity of snow as it hits the ground

Substituting appropriate values,

m(9.8)(10) + (1/2)(m)(4.43^2) = (1/2)(m)Vg^2

Since the factor "m" appears on both sides of the equation, it will simply cancel out, hence the above is modified to

(9.8)(10) + (1/2)(4.43^2) = (1/2)Vg^2

98 + 9.81 = (1/2)Vg^2

Vg^2 = 2(107.81) = 215.62

Vg = 14.68 m/sec.

Answer 2

NOTE: for ALL questions you MUST take into account the angle

a) for minimum value it must match the value of gravity (weight of object

at pitch of 30 degrees, gravitational acceleration is 9.8*sin 30=

4.9m/s^2 now force

F= 4.9m (let m be the mass from now on)

the normal force is 9.8* cos 30* m

now coefficient = F(static)/ F(normal)

9.8sin30m/9.8cos30m=

sin30/cos30= tan 30= 0.557350269 => minimum

b) we have to calculate the net acceleration of the snow chuck

NET acceleration= acceleration of gravity - acceleration caused by friction

acceleration of gravity= 9.8*sin 30= 4.9

force of kinetic friction= 0.2*9.8*cos 30*m

acceleration of friction= F/m

= 0.2*9.8cos 30m/m

= 0.2*9.8cos 30

net acceleration= 4.9-0.2*9.8cos 30= 3.202590209m/s^2

now find the time it takes to travel the distance

d= 0.5at^2

5= 0.5(3.202590209)*t^2

t= 1.767051936 s

now calculate its final speed

a= ( v(final)-v(initial) )/t  (inital speed was zero)

3.202590209= v(final)/1.767051936

v(final)= 5.65914323 m/s  => speed when it sildes off roof

c) since the roof is angled downwards then there will be 2 dimensions of motion, horizontal velocity V(hor) and vertical velocity V(ver)

V(hor)= 5.65914323*cos 30= 4.900961801m/s

V(ver)= 5.695914323*sin 30= 2.829571615m/s

now calculate time it takes to fall 10m

d=vt+0.5at^2

-10= -2.829571615t-0.5*9.8*t^2

solve for quadratic

t= 1.168725707

now its vertical speed would be

v=9.8*1.168725707+2.829571615

v=14.28308354m/s (vertical speed)

its final horizontal velocity is same throughout snow falling

4.900961801m/s

its final vertical speed is 14.280308354m/s

use pythagorean theorem for final velocity

so final velocity^ 2= vertical speed ^ 2+ horizontal velocity^2

v(final)^2= 14.28308354^2+4.900961801^2

v(final)= 15.1m/s => final velocity

hope this helps