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)

Question

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Answer 1

Let's set this up by listing every horizontal force acting on block Q directly, rather than analysing block P first.

Four horizontal forces act on Q: the applied force F pulling it forward, friction from the ground resisting that motion, friction from block P sitting on top of it, and the pull of the string, which loops from P over the fixed wall pulley and back to Q, so it also drags Q toward the wall.

Ground friction first. The ground carries the full weight resting on it, which is the weight of Q plus the weight of P transmitted down through it: $100+200=300\ \text{kN}$. With a friction coefficient of 0.3 between Q and the surface, this gives $0.3 \times 300=90\ \text{kN}$ of resistance.

Now consider what block P costs Q. Block P cannot move, the inextensible string holds it fixed against the wall. As Q is dragged out from under it, kinetic friction develops between P and Q equal to $0.4 \times 100=40\ \text{kN}$ (using the normal force of 100 kN, P's own weight). This entire 40 kN must be supplied by the string tension holding P still, so the tension in the string is also 40 kN.

Because the string loops back to Q through the same pulley, that same 40 kN tension pulls Q backward too, on top of the 40 kN friction reaction Q already feels from P. So block P effectively costs Q $40+40=80\ \text{kN}$ of resistance, not just 40 kN.

Adding the ground's contribution: $F=90+80=170\ \text{kN}$.

Let's summarize:

Ground friction on Q depends on the full weight of both blocks, giving 90 kN.
The friction between P and Q is 40 kN, and because the same string that resists this friction also loops back to Q, that 40 kN is felt twice.
Adding these gives the minimum pulling force.

So the minimum force needed to pull block Q is 170 kN.

Answer 2

Let's set this up by listing every horizontal force acting on block Q directly, rather than analysing block P first.

Four horizontal forces act on Q: the applied force F pulling it forward, friction from the ground resisting that motion, friction from block P sitting on top of it, and the pull of the string, which loops from P over the fixed wall pulley and back to Q, so it also drags Q toward the wall.

Ground friction first. The ground carries the full weight resting on it, which is the weight of Q plus the weight of P transmitted down through it: $100+200=300\ \text{kN}$. With a friction coefficient of 0.3 between Q and the surface, this gives $0.3 \times 300=90\ \text{kN}$ of resistance.

Now consider what block P costs Q. Block P cannot move, the inextensible string holds it fixed against the wall. As Q is dragged out from under it, kinetic friction develops between P and Q equal to $0.4 \times 100=40\ \text{kN}$ (using the normal force of 100 kN, P's own weight). This entire 40 kN must be supplied by the string tension holding P still, so the tension in the string is also 40 kN.

Because the string loops back to Q through the same pulley, that same 40 kN tension pulls Q backward too, on top of the 40 kN friction reaction Q already feels from P. So block P effectively costs Q $40+40=80\ \text{kN}$ of resistance, not just 40 kN.

Adding the ground's contribution: $F=90+80=170\ \text{kN}$.

Let's summarize:

Ground friction on Q depends on the full weight of both blocks, giving 90 kN.
The friction between P and Q is 40 kN, and because the same string that resists this friction also loops back to Q, that 40 kN is felt twice.
Adding these gives the minimum pulling force.

So the minimum force needed to pull block Q is 170 kN.

Show Hint

Correct Answer: 170

Solution and Explanation

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