A 10 kg block is pulled with a 50 N force at 30° above the horizontal (μ = 0.2). Find the acceleration — Physics problems | PhysLabo
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Acceleration of a block pulled at an angle

Medium 10 grade Mechanics / Newton's Laws
Problem statement

A block of mass $m = 10$ kg lies on a horizontal table. It is pulled by a force $F = 50$ N directed at an angle $\alpha = 30°$ above the horizontal (see the figure). The coefficient of friction between the block and the table is $\mu = 0.2$. Find the acceleration of the block. Take $g = 10$ m/s².

A block on a table pulled by force F at angle α above the horizontal; mg, N and friction are shown
Given
  • $m = 10$ kg
  • $F = 50$ N
  • $\alpha = 30°$
  • $\mu = 0.2$
  • $g = 10$ m/s²
  • $a = ?$
Hints
Hint 1 / 2
Draw all the forces and choose axes: one along the table, one vertical. Resolve $F$ into $F\cos\alpha$ and $F\sin\alpha$.
Hint 2 / 2
There is no vertical acceleration, so $N = mg - F\sin\alpha$ — the normal force is less than $mg$. Friction depends on exactly this $N$.
Solution

The forces on the block are gravity $m\vec g$, the normal force $\vec N$ from the table, the pulling force $\vec F$ and friction $\vec F_{fr}$. Write Newton's second law in components: $Ox$ along the table, $Oy$ vertically upward.

Projections of the forces onto the Ox and Oy axes
Projections of the forces onto the Ox and Oy axes

1. There is no vertical acceleration:

$$N + F\sin\alpha - mg = 0 \;\Rightarrow\; N = mg - F\sin\alpha = 100 - 50 \cdot 0.5 = 75 \text{ N}$$

The normal force is smaller than $mg$: part of $F$ lifts the block.

2. Kinetic friction:

$$F_{fr} = \mu N = 0.2 \cdot 75 = 15 \text{ N}$$

3. Horizontal direction:

$$ma = F\cos\alpha - F_{fr}$$$$a = \frac{F\cos\alpha - \mu\,(mg - F\sin\alpha)}{m} = \frac{50 \cdot 0.866 - 15}{10} = \frac{43.3 - 15}{10} \approx 2.83 \text{ m/s}^2$$
Answer

$a \approx 2.83$ m/s²

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