total mass = 2.5+1.2 = 3.7 kg
omega = 2 pi f
= 2pi/T = sqrt(2300/3.7) = 24.93
so
T = .252 second
normal force = m g
= 1.2 * 9.81 = 11.77 Newtons
max friction force = .56*11.77 = 6.59
when does m a = 6.59 N?
6.59 = 1.2 a
a = 5.49 meters/s^2
magnitude of a = A omega^2
= A(24.93)^2
so
A = 5.49/(24.93)^2 = .0088 meters
0.88 centimeters
Could someone please help walk me through this problem I'm struggling with? Thanks in advance!
A brick with mass m1=2.5 kg can move without friction on a horizontal plane. The brick
is attached to a wall by an ideal spring with spring constant k=2300 N/m. A second brick
with mass m2=1.2 kg lies on top of the brick as shown in Fig. 4. The static coefficient of
friction between the bricks is µs=0.56.
a) Assuming that m2 does not slide on m1 calculate the period, T, of small angle
oscillations for this system.
b) Calculate the maximum amplitude of oscillation at which m2 can follow along
without beginning to slide on m1
1 answer
1 answer