Range of equilibrium forces | CAIE 9709 Mechanics Notes

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Match each endpoint to an impending motion

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The endpoints of an equilibrium range come from opposite impending motions. Keep the applied force direction fixed and reverse friction.

For the least upward force, the body is about to slip down, so friction acts up the slope at $\mu R$ .
For the greatest upward force, the body is about to slip up, so friction acts down the slope at $\mu R$ .

Key idea

The applied force does not reverse. The impending motion reverses, so the friction arrow reverses.

Find R once, then solve both cases

Resolve perpendicular to the surface to find $R$ .
Calculate the limiting magnitude $F_{\max}=\mu R$ .
Draw and solve the least-force case with friction opposing downward slipping.
Reverse friction and solve the greatest-force case.
Write the final answer as an inequality or interval.

Common mistake

One limiting diagram gives only one endpoint. If the question asks for least and greatest values, two friction directions must appear in the working.

Turn the two cases into one range

Worked example

A force holds a block on a rough slope

A 5 kg block is held in equilibrium on a plane inclined at 20° to the horizontal. The coefficient of friction is 0.3. A force $P$ acts up the slope. Find the range of possible values of $P$ .

R=50\cos20^\circ=46.98\text{ N}

F_{\max}=0.3R=14.095\text{ N}

At the least value of $P$ , friction acts up the slope:

P_{\min}+14.095=50\sin20^\circ

P_{\min}=3.01\text{ N}

At the greatest value of $P$ , friction acts down the slope:

P_{\max}=50\sin20^\circ+14.095

P_{\max}=31.2\text{ N}

Answer

3.01\le P\le31.2\text{ N}

Examiner note

Least-and-greatest equilibrium questions require both impending-motion cases; the friction direction changes between them. (9709/42/M/J/24 Q5)

Your turnIndependent check [5 marks]

A 4 kg block is held in equilibrium on a plane inclined at 30° to the horizontal. The coefficient of friction is 0.2. A force $P$ acts up the slope. Find the range of possible values of $P$ .

Show worked answer

R=40\cos30^\circ=34.641\text{ N}

F_{\max}=0.2R=6.928\text{ N}

P_{\min}+6.928=40\sin30^\circ

P_{\min}=13.1\text{ N}

P_{\max}=40\sin30^\circ+6.928

P_{\max}=26.9\text{ N}

Answer

13.1\le P\le26.9\text{ N}

Match each endpoint to an impending motion

Swipe left or right to see the whole diagram →

The endpoints of an equilibrium range come from opposite impending motions. Keep the applied force direction fixed and reverse friction.

For the least upward force, the body is about to slip down, so friction acts up the slope at $\mu R$ .
For the greatest upward force, the body is about to slip up, so friction acts down the slope at $\mu R$ .

Key idea

The applied force does not reverse. The impending motion reverses, so the friction arrow reverses.

Find R once, then solve both cases

Resolve perpendicular to the surface to find $R$ .
Calculate the limiting magnitude $F_{\max}=\mu R$ .
Draw and solve the least-force case with friction opposing downward slipping.
Reverse friction and solve the greatest-force case.
Write the final answer as an inequality or interval.

Common mistake

One limiting diagram gives only one endpoint. If the question asks for least and greatest values, two friction directions must appear in the working.

Turn the two cases into one range

Worked example

A force holds a block on a rough slope

A 5 kg block is held in equilibrium on a plane inclined at 20° to the horizontal. The coefficient of friction is 0.3. A force $P$ acts up the slope. Find the range of possible values of $P$ .

R=50\cos20^\circ=46.98\text{ N}

F_{\max}=0.3R=14.095\text{ N}

At the least value of $P$ , friction acts up the slope:

P_{\min}+14.095=50\sin20^\circ

P_{\min}=3.01\text{ N}

At the greatest value of $P$ , friction acts down the slope:

P_{\max}=50\sin20^\circ+14.095

P_{\max}=31.2\text{ N}

Answer

3.01\le P\le31.2\text{ N}

Examiner note

Least-and-greatest equilibrium questions require both impending-motion cases; the friction direction changes between them. (9709/42/M/J/24 Q5)

Your turnIndependent check [5 marks]

A 4 kg block is held in equilibrium on a plane inclined at 30° to the horizontal. The coefficient of friction is 0.2. A force $P$ acts up the slope. Find the range of possible values of $P$ .

Show worked answer

R=40\cos30^\circ=34.641\text{ N}

F_{\max}=0.2R=6.928\text{ N}

P_{\min}+6.928=40\sin30^\circ

P_{\min}=13.1\text{ N}

P_{\max}=40\sin30^\circ+6.928

P_{\max}=26.9\text{ N}

Answer

13.1\le P\le26.9\text{ N}