What the Table Really Pushes With
A block of weight $40\ \text{N}$ rests on a horizontal table while a rope pulls it upwards at an angle $\theta$. The rope takes part of the load, so the table does not have to push with the full weight. Set the pull $P$ and the angle $\theta$ and watch the normal reaction $R$ change.
normal reaction $R = W - P\sin\theta$ = 30.0 N · horizontal pull $P\cos\theta$ = 17.3 N
The contact force from the table is not simply the weight: resolving vertically gives $R = W - P\sin\theta$.
Try each in your head before clicking an answer — this is a test, not a reveal. See how many you get first try. The clock starts on your first answer.
R1. The normal reaction from a surface acts:
R2. A block of weight $20\ \text{N}$ rests on a horizontal table with no other forces acting. The normal reaction is:
R3. A block resting on a table has zero resultant force. This is because:
When two surfaces touch, each surface exerts a single contact force on the other. It is convenient to split that force into two perpendicular parts:
| Part | Direction | Name |
|---|---|---|
| Normal component | perpendicular to the surface | normal reaction $R$ |
| Tangential component | along the surface | friction $F$ |
On a smooth surface the tangential part is zero, so the contact force is the normal reaction alone.
Always resolve perpendicular to the surface instead of quoting $mg$:
| Situation | Normal reaction |
|---|---|
| Block on a horizontal surface, no other vertical force | $R = W$ |
| Block on a smooth slope at angle $\theta$ | $R = W\cos\theta$ |
| Block pulled upwards at angle $\theta$ on a table | $R = W - P\sin\theta$ |
| Block pushed downwards at angle $\theta$ on a table | $R = W + P\sin\theta$ |
A surface is called smooth when there is no friction: the contact force has no component along the surface, so it acts perpendicular to the surface.
It isolates the normal reaction, so the geometry of resolving is easier to see before friction is added.
Every real surface is slightly rough. On a smooth slope a block would always slide; in practice it often stays put, which shows a friction component is present. A smooth model is an idealisation, not a fact.
Groups of 4 run a three-phase lab. Phase 1: move both sliders freely and describe the three arrows. Phase 2: fix W and increase theta — what happens to the normal reaction? Phase 3: answer the auto-graded challenge on contact forces and the smooth model.
Read each claim about a contact force and decide whether it is correct or contains an error, then read the explanation. Think about the normal reaction on a slope and the smooth model before you answer.
A block of weight $40\ \text{N}$ rests on a smooth plane inclined at $25^\circ$ to the horizontal. Find the normal reaction and the component of the weight acting along the plane.
Resolve the weight parallel and perpendicular to the plane. The perpendicular component is $W\cos 25^\circ$ and the parallel component is $W\sin 25^\circ$.
The plane is smooth, so the contact force is just the normal reaction. Resolving perpendicular to the plane: $R = W\cos 25^\circ = 40\times 0.9063 = 36.3\ \text{N}$.
The downhill component of the weight is $W\sin 25^\circ = 40\times 0.4226 = 16.9\ \text{N}$.
The normal reaction $36.3\ \text{N}$ is smaller than the weight. On a smooth plane nothing balances the $16.9\ \text{N}$ downhill component, so the block would accelerate — the model is not realistic here.
A block of weight $40\ \text{N}$ rests on a horizontal table. A rope pulls it with a force of $20\ \text{N}$ at $30^\circ$ above the horizontal. Find the normal reaction and the horizontal component of the pull. State what the smooth model assumes here.
The pull has a vertical component $20\sin 30^\circ$ upwards, which helps to support the block.
$R + 20\sin 30^\circ = 40$, so $R = 40 - 10 = 30\ \text{N}$.
$P\cos 30^\circ = 20\times 0.8660 = 17.3\ \text{N}$.
A smooth table has no friction, so nothing balances the $17.3\ \text{N}$ horizontal pull and the block would slide. In reality the table is rough and friction supplies the missing $17.3\ \text{N}$ along the surface.
Answer, then check. Options are shuffled every load. The clock starts on your first answer.
Q1. The contact force between two surfaces can be split into:
Q2. A block of weight $60\ \text{N}$ rests on a smooth slope at $40^\circ$ to the horizontal. The normal reaction is:
Q3. The smooth model of a surface assumes that:
Q4. A block of weight $50\ \text{N}$ rests on a horizontal table. A rope pulls it at $20^\circ$ above the horizontal with a force of $30\ \text{N}$. The normal reaction is:
Work each on paper, then reveal the solution to check your working.
A block of weight $30\ \text{N}$ rests on a smooth plane inclined at $35^\circ$ to the horizontal. Find the normal reaction.
Reveal Solution
$R = 30\cos 35^\circ = 24.6\ \text{N}$, perpendicular to the plane.
A block of weight $80\ \text{N}$ rests on a horizontal table. A rope pulls it at $25^\circ$ above the horizontal with a force of $25\ \text{N}$. Find the normal reaction.
Reveal Solution
$R = 80 - 25\sin 25^\circ = 80 - 10.6 = 69.4\ \text{N}$.
A sledge of mass $8\ \text{kg}$ is pulled at steady speed across level ground by a rope making $30^\circ$ with the horizontal. The resistance to motion is $10\ \text{N}$. Find the tension in the rope and the normal reaction.
Reveal Solution
Steady speed, so horizontally $T\cos 30^\circ = 10$, giving $T = 11.5\ \text{N}$. Vertically $R = 8\times 9.8 - 11.5\sin 30^\circ = 78.4 - 5.8 = 72.6\ \text{N}$.
A cyclist and cycle have total mass $67\ \text{kg}$. The greatest forward force she can produce is $200\ \text{N}$ and resistance totals $50\ \text{N}$. Find the angle of the steepest slope she can ride up at steady speed.
Reveal Solution
At steady speed the forward force balances resistance and the downhill component of the weight: $200 = 50 + 67\times 9.8\sin\theta$. So $\sin\theta = \dfrac{150}{656.6} = 0.2285$ and $\theta = 13.2^\circ$.
A block of weight $100\ \text{N}$ rests on a plane inclined at $20^\circ$ to the horizontal. (i) The plane is smooth — find the normal reaction. (ii) The plane is instead rough and the block is on the point of slipping down it. Describe how the contact force now differs from part (i).
Reveal Solution
(i) $R = 100\cos 20^\circ = 94.0\ \text{N}$ perpendicular to the plane.
(ii) The normal reaction is still $94.0\ \text{N}$, but the contact force
now also has a friction component of magnitude $\mu R$ acting up the slope.
The contact force is therefore no longer perpendicular to the plane; its magnitude is
$\sqrt{R^{2}+F^{2}}$ and it is tilted away from the normal.
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