Mathematics 9709/42 — May/June 2013

Cambridge A-Level · Mechanics · worked solutions for every part, with the mark scheme

questions

marks

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Topics Newton's Laws of Motion · Kinematics of Motion in a Straight Line · Forces and Equilibrium · Energy, Work and Power

Q1Forces and EquilibriumNewton's Laws of MotionFree sample

A string is attached to a block of weight $30\text{ N}$ , which is in contact with a rough horizontal plane. When the string is horizontal and the tension in it is $24\text{ N}$ , the block is in limiting equilibrium.

(i)

Find the coefficient of friction between the block and the plane.

DifficultyMedium-Easy

Worked solution

Approach

The plane is horizontal, so the normal reaction equals the weight. In limiting equilibrium the friction reaches its limiting value and balances the horizontal tension. Use $F=\mu R$ to find the coefficient.

Working

Resolving vertically:

R = 30\text{ N}

Since the string is horizontal and the block is in limiting equilibrium,

F = T = 24\text{ N}

Using $F=\mu R$ :

\mu = \frac{F}{R} = \frac{24}{30} = 0.8

Answer

\mu = 0.8

Final answer

0.8

Detailed explanation

Walkthrough

Since the plane is horizontal, the normal reaction $R$ is the force pressing the block onto the plane; with no vertical acceleration it must balance the weight, so $R=30\text{ N}$ . The block is in limiting equilibrium, which means the friction has its greatest possible value and is exactly balancing the horizontal tension $T=24\text{ N}$ . The limiting friction law is $F=\mu R$ . Substitute the values and solve for $\mu$ . This is a direct application of the force balances in two perpendicular directions.

Key Takeaways

On a horizontal plane with no vertical acceleration, $R=W$ .
In limiting equilibrium the friction force satisfies $F=\mu R$ .
The coefficient of friction is the dimensionless ratio $\frac{F}{R}$ .

Common Mistakes

Applying $F=\mu R$ without first identifying the normal reaction.
Assuming friction equals the tension without recognizing the block is in limiting equilibrium.
Mixing up $F$ and $R$ when dividing: the coefficient is $F/R$ , not $R/F$ .

Things to Be Careful About

Here $R=W=30\text{ N}$ only because the plane is horizontal and there is no extra vertical force. This changes in part (ii).
The mark scheme requires the method $R=W$ , $F=T$ and $F=\mu R$ ; an unsupported answer is not sufficient.

Techniques used

resolve vertical forces to find the normal reactionuse the equilibrium condition horizontallyapply the limiting friction lawsolve for the coefficient of friction

(ii)

The block is now in motion and the string is at an angle of $30^\circ$ upwards from the plane. The tension in the string is $25\text{ N}$ .

Find the acceleration of the block.

DifficultyMedium

Worked solution

Approach

Resolve vertically to find the normal reaction when the string is inclined, then use $F=\mu R$ to find the friction. Resolve horizontally and apply Newton's second law to find the acceleration. Take $g=10\text{ ms}^{-2}$ .

Working

With upward vertical as positive, the vertical component of the tension is $25\sin 30^\circ$ , so:

R + 25\sin 30^\circ = 30

Therefore

R = 30 - 25 \times \frac12 = 17.5\text{ N}

The frictional force is

F = \mu R = 0.8 \times 17.5 = 14\text{ N}

The horizontal component of the tension is $25\cos 30^\circ$ . Applying Newton's second law horizontally:

25\cos 30^\circ - F = ma

The mass is

m = \frac{30}{g} = \frac{30}{10} = 3\text{ kg}

Therefore

25\cos 30^\circ - 14 = 3a

Using $\cos 30^\circ = \frac{\sqrt{3}}{2}$ :

25 \times \frac{\sqrt{3}}{2} - 14 = 3a

Evaluating:

21.65 - 14 = 3a

3a = 7.65

a = 2.55\text{ ms}^{-2}

Answer

2.55\text{ ms}^{-2}

Final answer

2.55 ms^-2

Detailed explanation

Walkthrough

Now the string is inclined at $30^\circ$ upwards, so its tension has a horizontal component $25\cos 30^\circ$ and a vertical component $25\sin 30^\circ$ . The vertical component acts upwards and therefore reduces the normal reaction from the plane. There is no vertical acceleration, so the upward forces equal the downward weight: $R+25\sin30^\circ=30$ . This gives $R=17.5\text{ N}$ . Using the coefficient found in part (i), $F=\mu R=0.8(17.5)=14\text{ N}$ . Horizontally, the net force is $25\cos30^\circ-14$ . Newton's second law says this equals mass times acceleration. The mass is found from $W=mg$ , so $m=30/g=3\text{ kg}$ with $g=10\text{ ms}^{-2}$ . Solve $25\cos30^\circ-14=3a$ to get $a=2.55\text{ ms}^{-2}$ .

Key Takeaways

An inclined force must be resolved into components.
An upward component of a force reduces the normal reaction, which in turn reduces friction.
The friction force during motion is still found using $F=\mu R$ once $\mu$ is known.
Newton's second law is applied to the horizontal direction because the acceleration is horizontal.

Common Mistakes

Using $R=30\text{ N}$ for friction and forgetting the vertical component of tension; the mark scheme requires $R=30-25\sin30^\circ$ .
Using $25\sin 30^\circ$ as the horizontal component instead of $25\cos 30^\circ$ .
Using the weight $30\text{ N}$ as the mass in $F=ma$ ; the mass is $30/g$ , not $30$ .
Omitting the $a$ or using a value of $g$ other than $10$ where the mark scheme uses $g=10$ to obtain $2.55$ .

Things to Be Careful About

The block is moving, so it is no longer in limiting equilibrium, but the value of $\mu$ from part (i) is still used.
Friction opposes the horizontal motion, so it is subtracted from the horizontal component of tension.
Take care to use the vertical equilibrium equation $R+T\sin30^\circ=W$ , not $R=W$ .
The final acceleration has units $\text{ms}^{-2}$ ; if a different value of $g$ is used, the rounded answer may differ slightly.

Techniques used

resolve the inclined tension into componentsuse the friction law with the reduced normal reactionapply Newton's second law horizontallyconvert weight to mass and solve for acceleration

The rest of this paper

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