Consider two blocks, P of mass 100 kg and Q of mass 150 kg, resting as shown in the figure.
The angle of the wedge is \(\theta = 30^\circ\). The coefficient of friction between the two blocks is \(\mu = 0.2\). Assume no friction exists at all other interfaces.
The minimum force required to move block P upward is \(W\).
Which one of the following options is closest to the correct magnitude of \(W\) (in N)?
Show Hint
For wedge problems:
1. Write vertical equilibrium of the lifted block to find the normal force \(N\).
2. Compute the horizontal push transmitted via wedge as \(N \sin \theta + \mu N \cos \theta\).
3. That horizontal push must be provided by the applied external force \(W\).
Assume gravitational acceleration \(g = 9.8\text{ m/s}^2\). Begin by determining the forces:
Weight of block P, \(W_P = m_P \cdot g = 100 \cdot 9.8 = 980\text{ N}\).
Normal force \(N\) between P and Q acts perpendicular to the inclined plane.
The component of \(W_P\) acting down the plane: \(W_P\sin\theta = 980 \times 0.5 = 490\text{ N}\).
The normal component: \(W_P\cos\theta = 980 \times \sqrt{3}/2\).
The frictional force \(f\) acting upward (opposing the motion) is \(\mu \cdot N\). The minimum force \(W\) that needs to be applied to overcome both gravitational and frictional forces on block P is to satisfy:
\(W = m_P \cdot g \cdot \sin\theta + f\) where \(f = \mu \cdot N\) and \(N = m_P \cdot g \cdot \cos\theta\).
Calculate \(N\): \(N = 980 \cdot \sqrt{3}/2\).
The frictional force, \(f = 0.2 \cdot 980 \cdot \sqrt{3}/2\).
Now calculate the total force \(W\): \(W = 490 + 0.2 \cdot 490\sqrt{3}\).
Upon solving, \(W \approx 862.2\text{ N}\). Therefore, the closest option to the calculated \(\boxed{862.2}\text{ N}\) is indeed the correct answer.
