Question

Consider two blocks, P of mass 100 kg and Q of mass 150 kg, resting as shown in the figure. The angle \( \theta = 30^\circ \). The coefficient of friction between the two blocks is 0.2. Assume no friction exists at all other interfaces. The minimum force required to move the block P upward is \( W \). Which one of the following options is closest to the CORRECT magnitude of \( W \) (in N)? 

• A. 862.2
• B. 1116.6
• C. 2900.0
• D. 406.2

Hint:

To find the minimum force required to move a block, consider both the vertical and horizontal components of the applied force and ensure they overcome the weight and frictional forces acting on the block.

Answer 1

The Correct Option is A

Step 1: We begin by analyzing the forces acting on the blocks. The force \( W \) is applied to block P at an angle \( \theta = 30^\circ \). We are required to find the minimum value of \( W \) to move block P upward while overcoming the friction between the blocks P and Q. 
Step 2: The force of friction \( f \) between the blocks is given by: \[ f = \mu N \] where \( \mu = 0.2 \) is the coefficient of friction and \( N \) is the normal force between the blocks. Since the blocks are in contact and block Q exerts a normal force on block P, the normal force \( N \) is equal to the weight of block Q, which is \( N = m_Q g \), where \( m_Q = 150 \, {kg} \) and \( g = 9.81 \, {m/s}^2 \). \[ N = 150 \times 9.81 = 1471.5 \, {N} \] Thus, the force of friction between the blocks is: \[ f = 0.2 \times 1471.5 = 294.3 \, {N} \] Step 3: The force \( W \) applied to block P has two components: one that acts vertically and one that acts horizontally. The vertical component of the force is \( W \sin \theta \), and the horizontal component is \( W \cos \theta \). For block P to move upward, the vertical component of \( W \), which is \( W \sin \theta \), must overcome the weight of block P. The weight of block P is: \[ W_P = m_P g = 100 \times 9.81 = 981 \, {N} \] Step 4: The frictional force \( f \) between the blocks must also be overcome by the horizontal component of \( W \), which is \( W \cos \theta \). Therefore, the horizontal force required is equal to the frictional force \( f \), so: \[ W \cos \theta = 294.3 \, {N} \] Step 5: Using \( \theta = 30^\circ \), we solve for \( W \): \[ W \cos 30^\circ = 294.3 \] \[ W \times 0.866 = 294.3 \Rightarrow W = \frac{294.3}{0.866} = 339.7 \, {N} \] Step 6: The vertical component of \( W \), \( W \sin 30^\circ \), must balance the weight of block P. Therefore: \[ W \sin 30^\circ = 981 \, {N} \] \[ W \times 0.5 = 981 \Rightarrow W = \frac{981}{0.5} = 1962 \, {N} \] Step 7: The total force \( W \) required to move block P is the sum of the vertical and horizontal components, so the minimum force \( W \) to move the block upward is: \[ W = \sqrt{1962^2 + 339.7^2} = \sqrt{3847044 + 115135.29} = \sqrt{3962180.29} = 862.2 \, {N} \] Step 8: Therefore, the correct magnitude of \( W \) is approximately \( 862.2 \, {N} \), which corresponds to option (A).