# Actual mass and Apparent mass due to Force

Suppose a block is placed on a surface (with a non-negligible frictional constant) and a person exerts a constant force $F_1$ on it parallel to the surface. Whether the force is able to cause displacement or not depends on the mass $m$. The force and mass are related as $F = ma$. Suppose this force causes displacement.

Now if a second force $F_2$ constantly acts on the block vertically downwards such that it comes to a rest provided $F_1$ still acts on the body, the person perceives the mass of the body to have increased while no actual change can occur in its mass.

How would you calculate the apparent increase in the mass of the body, the apparent mass?

Note: The real mass of the body cannot change but the perceived or apparent mass does. Addition of further assumptions is acceptable. The above problem takes friction into account.

Whether the force is able to cause displacement or not depends on the mass m.

This is incorrect. A force will always attempt to accelerate an object, regardless of its mass, which causes displacement. The real factor here is whether friction is present. If the applied force is smaller than the static friction, the block will not move. If the applied force is greater than the static friction, the block will accelerate with an acceleration less than the applied force as friction in the form of kinetic friction will oppose the applied force.

Now if a second force F2 acts on the block vertically downwards such that it comes to a rest

Again, this is incorrect. The block cannot come to a rest if the two forces are the only forces acting on the block. When $F_2$ is applied, it will be opposed by an equal and opposite normal force by the surface. I think that you are again talking about friction. The formula for friction is $$F = \mu N$$ where $\mu$ is the coefficient of friction and $N$ is the normal force. So by pressing down on the block, you are increasing the friction, presumably until it has a magnitude greater than $F_1$, causing an acceleration in the opposite direction, until it comes to a stop.

person perceives the mass of the body to have increased

The mass of a body is an intrinsic quantity, and always remains constant. You are confusing mass with weight. Weight is the gravitational force acting on the block. Since the weight of the block is perpendicular to the acceleration, I do not see why either quantity will increase. The only quantity that changes is the total downward force acting on the block which is simply the sum of the weight and $F_2$. This is, of course, opposed by an equal and opposite normal force from the surface.

• Yes, friction is present. I will try to clarify the problem later. – Muhammad Inam Sep 9 '18 at 13:23

There is no apparent increase in mass. The acceleration is always given by $\Sigma F=ma$ where $\Sigma F$ is called the net force.

Often Newton’s 2nd law is written as $F=ma$ for brevity or with the understanding that there is only one force. However, this can lead to the confusion that you are encountering here.

The correct expression indicates that you must consider all forces acting on the object, not just one. When you do that you find that $\Sigma F=ma$ with the same mass in all cases.

In both cases the vertical forces cancel out and the horizontal forces are the applied force and a frictional force. Assuming that the block is moving in the positive direction and that the applied force is also in the positive direction then $\Sigma F=F_{applied}-F_{friction}=ma$ in both cases, with the only difference being the magnitude of the friction force is larger in the second case.

How would you calculate the apparent increase in the mass of the body, the apparent mass?

If you are pushing a block along a surface with friction, its acceleration will be lower than it would be without friction, i.e., $a'=\frac {F-f} m < a=\frac F m$.

We can say that, if you assume that there is no friction, i.e., that the only force acting on the block is $F$, the perceived mass of the block, in the presence of friction, would be greater than its actual mass $m'=\frac F {a'} > m=\frac F a$.

Of course, you can roughly determine the friction by gradually increasing the applied force and marking the force level at which the object starts to move.

• The problem with this is that the relation F=m’a does not hold. If you double F you do not double a, and for small F you do not get small a but a=0 instead. There is no sense in which this apparent mass works. – Dale Sep 9 '18 at 14:50
• @Dale It works for a fixed applied force exceeding the force of friction, which is what PO was asking about. Of course, you can quickly determine that the relationship is not linear by changing the force. This is how I have proposed to figure out the value of the friction force. – V.F. Sep 9 '18 at 15:05