17.9 kg crate slides from rest without friction 4.2m down a ramp incllned 17.6 o
ID: 1277083 • Letter: 1
Question
17.9 kg crate slides from rest without friction 4.2m down a ramp incllned 17.6o below horizontal. in the two cases below, the same calculations will be performed using different coordinate system
1. calculate displacement, normal force, acceleration and velocity vectors at the ramp's end using coordinates that are parallel (downwards) and perpendicular (above) to the ramp
2. calculate displacement, normal force, acceleration and velocity vectors at the ramp's end using vertical and horizontal (standard) coordinates.
Explanation / Answer
1.
displacement:
parallel = 4.2, perp = 0
normal force:
sum forces perp
N - m g cos theta = 0
N = m g cos theta = 17.9*9.81*cos(17.6 degrees)= 167.4 N
acceleation:
sum forces parallel
m g sin theta = m a
a = g sin theta = 9.81*sin(17.6 degrees)= 2.97 m/s^2
so (2.97 m/s^2, 0)
velocity:
at the end v^2 = v0^2 + 2 a x
v = sqrt(2*2.97*4.2)= 4.99 m/s
so vector is (4.99 m/s, 0)
2. now use standard x and y
displacemet:
x displacement = 4.2*cos(17.6)=4.0 m
y displacement = -4.2*sin(17.6)=-1.27 m
normal force:
sum forces in the y
N cos theta - mg = m ay
in the x
N sin theta = m ax
use that ax = a cos theta
ay = -a sin theta
so
N sin theta = m a cos theta
a = N tan theta/m
plug into y
N cos theta - mg = -N tan theta sin theta
multiply both sides by cos theta
N cos theta^2 - m g cos theta = -N sin theta^2
N cos theta^2 + N sin theta^2 = m g cos theta
cos^2 + sin^2 = 1
so N = m g cos theta = 167.4 N
acceleration vectors:
a = 167.4*tan(17.6 degrees)/17.9= 2.97 m/s^2
so a vector = (2.97*cos(17.6 degrees), -2.97*sin(17.6 degrees))
=(2.83 m/s^2, -0.898 m/s^2)
velocity vectors:
use v^2 = v0^2 + 2 a d
so in the x, v = sqrt(2*2.83*4)= 4.76 m/s
in the y, v = -sqrt(2*-0.898*-1.27)= -1.51 m/s
so velocity vector (4.76 m/s, -1.51 m/s)
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