Suppose a block with mass $3\ \mathrm{kg}$ is on a frictionless table and a force of $15\ \mathrm N$ pulls it from an angle of $38^\circ$ above the horizontal.

1) What is the apparent weight of the block?

2) What is the minimum force for the block to take off?

Solution Given

$$\sum F_y = n - mg - F\sin\theta = 0$$

$$\sum F_x = F\cos\theta = ma_x$$

Solve for $n$, the apparent weight or the normal force.

$$n = 20.16\ \mathrm N$$

The minimum force to take off when $n = 0$

$$mg = F\sin\theta$$

$$F = \frac{mg}{\sin\theta}$$

$$F = 47.75\ \mathrm N$$


Why on earth is the force on the vertical direction $0$? Shouldn't it be

$$\sum F_y = n - mg - F\sin\theta = ma_y$$

There is a horizontal acceleration component, but no vertical acceleration.


3 Answers 3


You're correct that the vertical component of Newton's second law should be

$$\sum F_y = ma_y$$

You set $a_y = 0$ because the block is not flying up off the table. This is implied by the wording of the problem: usually blocks are assumed not to be flying up off their tables unless it is explicitly stated that that is a possibility. Plus, the fact that they ask for an apparent weight suggests that the block is staying against the surface it sits on.

However, you can also show that the block stays on the table if the diagonal force is $15\text{ N}$. To do so, you use the rule that the normal force will be as strong as it has to be to cancel out the forces pushing the block against the surface.

In detail, you add up all forces acting on the block other than the normal force, which in this case gives you $F\sin\theta - mg$. If this "subtotal force" is directed into the surface, the normal force will have the same magnitude but will act away from the surface, so that the net force including the normal force is zero.

$$N + \sum_\text{other} F_\perp = 0$$

On the other hand, if the "subtotal force" is directed away from the surface, the normal force will be unable to counteract it. In that case the normal force will be zero, and there will be a net force on the block. Since net force equals acceleration, you can then conclude that the block will have an acceleration perpendicular to the table, i.e. it will fly up off the table.

In short, the status of the block's motion and the normal force depends on the sum of the other forces:

  • $\sum_\text{other} F_\perp$ acts toward surface: $N > 0$, no motion
  • $\sum_\text{other} F_\perp = 0$: $N = 0$, no motion
  • $\sum_\text{other} F_\perp$ acts away from surface: $N = 0$, motion away from surface
  • $\begingroup$ But isn't the 15N force "pulling" the block "away"? Doesn't that fall under condition three? $\endgroup$
    – Lemon
    Commented Dec 18, 2011 at 20:30
  • $\begingroup$ The conditions are applied only to the sum of all forces other than the normal force. That's why the sum sign is in there. In particular, don't forget about gravity. $\endgroup$
    – David Z
    Commented Dec 19, 2011 at 1:24

The vertical acceleration will be less than the acceleration due to gravity, hence the block will stay on table. For this reason the vertical acceleration is taken to be zero.


Even if the reading the conditions of the problem and question 1) didn't clue you on the movement of the block, question 2) clearly indicates that the block is resting on the table since it asks what it would take to lift it off. As usual, when going through a new problem, make sure you read all questions first as they might clue you on the kind of assumptions you can make to solve the problem.


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