A block of mass m1 = 5.0 kg is at rest on a plane that makes an angle = 30.0° ab

Question

A block of mass m1 = 5.0 kg is at rest on a plane that makes an angle = 30.0° above the horizontal. The coefficient of static friction between the block and the incline is 0.47. The block is attached to a second block of mass m2 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 (m2,max)

kg (m2,min)

(b) What is the magnitude of the frictional force on the 5.0 kg block if m2 = 1.0 kg?

N

Explanation / Answer

uk = 0.47
m1 = 5.0 kg
(a)
Calculatming m2 max,

T = m1*g*sin(30) + uk*m1*g*cos(30)
T = 5.0 * 9.8 * sin(30) + 0.47 * 5.0 * 9.8 * cos(30)
T = 44.44 N

T = m2 max*g
m2 max = 44.44/9.8
m2 max = 4.53 kg

Calculatming m2 min,
T = m1*g*sin(30) - uk*m1*g*cos(30)
T = 5.0 * 9.8 * sin(30) - 0.47 * 5.0 * 9.8 * cos(30)
T = 4.55 N

T = m2 min*g
m2 min = 4.55/9.8
m2 min = 0.464 kg

Range : - < 0.464 , 4.53>

(b)
T = 1.0 * 9.8 N = 9.8 N

m1*g*sin(30) = 5.0 * 9.8 * sin(30) = 24.5 N

Friction Force,
24.5 - T = Friction Force
Friction Force, = 24.5 - 9.8
Friction Force, = 14.7 N

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