A block of mass m 1 = 4.0 kg is at rest on a plane that makes an angle = 30.0° a

Question

A block of mass m1 = 4.0 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.

(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 4.0 kg block if m2 = 1.0 kg?
_______ N

Explanation / Answer

part a )

First, the frictional force:

Ff = mu*N where mu is the coefficient of static friction

N = m1gcostheta

theta = 30degree

So, Ff = mu*m1*gcostheta

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)

(i) Minimum m2:

The equations are:

T = m2g...(i)

m1gsintheta = T + frictional force opposing the downward motion

or m1gsintheta = T + mu*m1gcostheta ...(ii)

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

m1gsintheta = m2g + mu*m1gcostheta

or m1sintheta = m2 + mu*m1*costheta

m2 = m1sintheta - mu*m1costheta

or m2 = m1(sintheta - mu*costheta)


m2(min) = 0.545 kg

(ii) 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

m1gsintheta+ mu*m1*gcostheta = T....(iii)

:friction is opposing the upward motion of m1

Substitute for T in (iii) to get

m1gsintheta + mu*m1gcostheta = m2g

or m1sintheta + mu*m1costheta = m2

or m2 = m1(sintheta + mu*costheta)

m2 = 3.45 kg

3.45 kg max

0.545 kg min

part b )

The frictional force on the 4 kg block if m2 = 1.0 kg

Let F be the friction

m1gsintheta = T + F

T = m2g

So m1gsintheta = m2g + F

F = m1gsintheta - m2g

F = g(m1sintheta - m2)

F = 9.8 N

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