lab manual 3328-Adv D EEE3221 lab Manual Figure 1. Astable Mutt x A Google trans
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lab manual 3328-Adv D EEE3221 lab Manual Figure 1. Astable Mutt x A Google transl x U1 Quiz Chap os x C search Q&A; lchegg x C O www.webassign.net/web/Student/Assignment Responses ast?dep 154 1025 A block of mass m1 6.0 kg is at rest on a plane that makes an 0 30.0° above the horizontal. The coefficient of static friction between the block and the incline is 047. The block is attached to a second block of mass m2 angle s that hangs freely by a string that passes over a frictionless and massless pulley (a) Find the range of possible values for m2 for which the system will be in static equilibrium. kg (m max kg (m2 min) (b) What is the magnitude of the frictional force on the 6.0 kg block if m2 1.0 kg? submit Answer save Progress My Notes o Ask your 2 O 1 points Tipler6 5.P 047 soln A 150-g block is projected up a ramp with an initial speed of 7.2 m/s. The coefficient of kinetic friction between the ramp and the block is 0.23. (a) If the ramp is inclined 30° with the horizontal, how far along the surface of the ramp does the block slide before coming to a stop? (b) What minimum coefficient of static friction between the block and the ramp if the block is not to slide back down the ramp? is the 3. o -5.5 points Tipier6 SP 069Explanation / Answer
GIVEN
mass m1 = 6 kg, m2 = ?
theta = 30 degrees
coefficient of static friction mue_s = 0.47
the frictional force is f = mue_s*mg cos theta =
the minimum value of mass m2 is if m1 move down and for max m2 , m1 should move upward
say for minimum value of m2
her the force on m2 is T -m2g = 0 ==> T = m2g
m1g sin theta = T+ mue_s m1g cos theta
m1g sin theta = m2*g+ mue_s m1g cos theta
m2 = m1 (sin theta - mue_s cos theta)
= 6(sin30-0.47cos30) kg
= 0.55781 kg
for maximum value of m2
T = m2*g
and
m1g sin theta+mue_s*m1*g cos theta = T
m1 sin theta + mue_s m1 cos theta = m2
m2 = m1(sin theta + mue_s cos theta)
m2 = 6(sin30+0.47cos30)
m2 = 5.44219 kg
b) magnitude of the frictional force on m1 if m2 = 1 kg
m1g sin theta = T+f and T = m2g
m1g sin theta -m2g= f
f = g(m1sin theta-m2)
f = 9.8(6 sin30 - 1) N
f = 19.6 N
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