Question

Two blocks P and Q weighing \(100\ \text{kN}\) and \(200\ \text{kN}\), respectively, are connected by a string passing through a massless and frictionless pulley as shown. The friction coefficient between blocks P and Q is 0.4, and that between block Q and surface is 0.3. The minimum force, F, in kN, required to pull the block Q, is . (answer in integer)

Hint:

Notice the same string ties both blocks to the fixed pulley on the wall, so think about how the tension needed to hold block P also acts back on block Q.

Answer 1

Correct Answer: 170

Step 1: Understand the arrangement in the figure.
Block P (weight 100 kN) rests on top of block Q (weight 200 kN), which sits on the ground. A single string runs from block P to a pulley fixed on the wall and back to block Q, so the same tension T pulls both blocks toward the wall.

Step 2: Find the friction between P and Q, and hence the string tension.
The only horizontal forces on P are the string tension and the friction from Q sliding beneath it. The normal force between P and Q equals the weight of P, 100 kN, so this friction is \(0.4 \times 100=40\ \text{kN}\).
Since P stays in place (held in position by the string), this friction must be balanced entirely by the string tension, so \(T=40\ \text{kN}\). By Newton's third law, Q feels a 40 kN friction reaction from P, opposing its motion.

Step 3: Find the friction between Q and the ground.
The ground supports the weight of both blocks, since P's weight is transmitted down through Q. Normal force on the ground \(=100+200=300\ \text{kN}\).
Friction between Q and the ground \(=0.3 \times 300=90\ \text{kN}\).

Step 4: Add up every force resisting F.
To pull Q, F must overcome: the 40 kN friction reaction from P, the 90 kN friction from the ground, and the 40 kN string tension, which also pulls Q back toward the wall through the same pulley.
\(F=40+90+40=170\ \text{kN}\).

Final Answer:
The minimum force needed to pull block Q is \(170\ \text{kN}\). \[ \boxed{170} \]