Question

What is the minimum weight (W1) required to hold the block (M2) static on the ramp, if the ramp is inclined 30 degrees and the coefficient of friction between the block and the ramp is 0.5? Assume the pulley is frictionless.

Answer 1

The minimum weight
`(\(W_1\))`required to hold the block
`(\(M_2\))` static on the ramp, inclined at 30 degrees with a coefficient of friction (\(\mu\)) of 0.5, is
`\(W_1 = (M_2 \cdot g)/(\sin(30^\circ) + \mu \cdot \cos(30^\circ))\ ).`

Step-by-step explanation:

To determine the minimum weight required
`(\(W_1\)),` we can use the equation for static equilibrium in the direction perpendicular to the inclined plane. The gravitational force component parallel to the ramp is given by
`\(M_2 \cdot g \cdot \sin(30^\circ)\),` while the friction force opposing motion is
`\(\mu \cdot N \cdot \cos(30^\circ)\)`, where
`\(N\)`is the normal force. At the minimum weight
`(\(W_1\)),` the net force in this direction is zero. Thus, we can set up the equation:

`\[M_2 \cdot g \cdot \sin(30^\circ) = \mu \cdot N \cdot \cos(30^\circ)\]`

The normal force (\(N\)) is equal to the gravitational force component perpendicular to the ramp, which is
`\(M_2 \cdot g \cdot \cos(30^\circ)\).`Substituting this into the equation, we get:

`\[M_2 \cdot g \cdot \sin(30^\circ) = \mu \cdot M_2 \cdot g \cdot \cos(30^\circ)\]`

Now, solving for
`\(W_1\)`:

`\[W_1 = (M_2 \cdot g)/(\sin(30^\circ) + \mu \cdot \cos(30^\circ))\]`

This formula gives the minimum weight
`(\(W_1\))` required to hold the block static on the inclined ramp. The calculation involves the gravitational constant
`(\(g\)),` the mass of the block
`(\(M_2\)),` the angle of inclination (30 degrees), and the coefficient of friction
`(\(\mu\))`