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

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 0.42. 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 30 (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 m21.0 kg? eBook

Explanation / Answer

(a) For mass m1, the frictional force is -  

Ff = µN where µ is the coefficient of static friction

N = m1gcos

= 30°

So, Ff = µm1gcos

Now, to find the range of values for m2, we will look at the extreme values
that will just allow the mass m1 to move down (min m2) and the value
of m2 that will just allow the mass m1 to move up (max m2)

(A) Minimum m2:

The equations are:

T = m2g...(i)

m1gsin = T + frictional force opposing the downward motion

or m1gsin = T + µm1gcos...(ii)

Substitute eqn (i) into eqn (ii). Then

m1gsin = m2g + µm1gcos

or m1sin = m2 + µm1cos

Thus m2 = m1sin - µm1cos

or m2 = m1(sin - µcos)

Compute this to get

m2(min) = 5*(sin30 - 0.42*cos30) = 5*(0.50 - 0.364) = 0.68 kg.

(B) Maximum m2:

We want to find m2 that will just be enough mass to start m1
moving up!

The equations become

T = m2g again, and

m1gsin + µm1gcos = T....(iii)

:friction is opposing the upward motion of m1

Substitute for T in (iii) to get

m1gsin + µm1gcos = m2g

or m1sin + µm1cos = m2

or m2 = m1(sin + µcos)

Compute m2 to get:

m2(max) = 5*(sin30 + 0.42*cos30) = 5*(0.5 + 0.50* 0.87) = 4.66 kg

The range for m2, therefore, is

4.66 kg m2 0.68 kg

(b) Consider F is the requisite frictional force.

m1gsin = T + F

T = m2g

=> m1gsin = m2g + F

=> F = m1gsin - m2g

=> F = g(m1sin - m2)

=> F = 9.81*(5*sin30 - 1.0) = 9.81* 1.5 = 14.72 N

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