When two bodies come in contact -surfaces, a component acts upon them perpendicular to the contact surface. This is known as normal force.
Can normal force be at an angle:- The normal force exerted on the body tends to balance the gravity force (mg) on the body. As the normal force depends upon the value of force experienced by a body, but the normal force is always perpendicular to the body.
We need to understand the force exerted on a body. Let us take an example of a block that is at rest position on the table surface.
In this condition, the block experiences two types of forces.
- First is the gravity force (mg) which acts vertically downwards at the center of gravity of the block.
- Second is the reactionary force P which acts vertically upward. These forces are passing through the center of gravity of the block.
Hence the block is in P= mg.
Now, if we apply any external force F on the block, suppose in the right direction. In this condition, the block does not move. Instead, the force P ( vertically upward) is now inclined in the left direction. Here the force P acting on the block can be divided into two components. One will be parallel, and the other will be perpendicular to the contact surface.
Force of static friction this force balance the applied force F . in contrast, the force which is perpendicular to the block is known as normal force R . (R=mg).
The limiting frictional force is directly proportional to the normal force:-
f_{s}αR
The coefficient of limiting friction is the ratio of limiting frictional force to the normal force.
When two surfaces are in relative motion, the force acting between them is known as kinetic frictional force fk. This is less than the limiting fraction, this is given as:-
fk=μkR
Where ,
μk<μs
To check whether “can normal force be at an angle” lets check it through the following conditions:-
While pulling roller:-
When a roller of mass m is tried to pull over a horizontal surface by applying a force of F at an angle . As shown below in the figure:-
By considering that roller is in equilibrium, we say that:-
R_{1}+Fcosθ=mg
R_{1}=mg-Fcosθ
Here the coefficient of static friction between roller and surface is, then we can write f_{s} as:-
f_{s}=μ_{s}R1
f_{s}=μ_{s}(mg-Fcosθ)
Now we can say that normal force is perpendicular to the center of gravity of the roller.
When a roller of mass m is tried to push over a horizontal surface by applying a force of F at an angle. As shown below in the figure:-
R2=Fcosθ+mg
f_{s}‘=μ_{s}R2
f_{s}‘=μ_{s}(Fcosθ+mg)
In this condition also the normal force is perpendicular to the center of gravity of the roller.
From the given an example, while comparing Equations 1 and 2, we can say that it is easier to pull a roller rather than to push it because frictional force is less while pulling.
If we consider a case of rolling friction which is much less than sliding friction, in this case, also we find normal force at 90 degrees only.
As we know, the rolling friction is directly proportional to the normal force R. and inversely proportional to the radius r of the wheel.
f_{r}αR/r
f_{r}=μ_{r}*(R/r)
Where μ_{r} is the coefficient of rolling friction.
From all the given examples, we know that normal force is always perpendicularly upward to the center of gravity of the body.
Problem examples related to normal force:-
- A force of 980N is just able to move a block of having a mass of 200kg on a rough horizontal; surface. Calculate the coefficient of friction and the angle of friction?
The force of 980 N is equal to the limiting frictional force. Hence the coefficient of static friction is:
μ_{s}=f_{s}/R
Where R is the normal force is equal to mg.
μ_{s}=980/(200*9.8)
μ_{s}=0.05
The angle of friction is given by;
tan θ_{s}=μ_{s}=0.05
θ_{s}=tan^{-1}(0.05)
- A block of mass 2kg is placed on the floor. The coefficient of static friction is 0.4. A force of 2.5 N is applied on the block, as shown. Calculate the force of friction between the block and the floor?
Let R be the normal force on the block exerted by the floor. The limiting force of static friction is:
f_{s}=μ_{r}R=μ_{s}mg
0.4*2kg*9.8ms^{2}
7.84N
- When a weight of a body placed on a surface is doubled, how does the coefficient of friction change?
There is No change in the coefficient of friction. In fact, the force of the limiting fraction is doubled.
f_{s}=μ_{s}R=μ_{s}mg
- Explain how lubricating can help in reducing friction?
When we lubricate a body, then the lubricant form a thin layer around two surface. In such condition the sliding friction is replaces by liquid friction. Where liquid friction is less than sliding friction. this results in less friction.
- Can we jump off from a frictionless horizontal surface?
No, we can not jump off from a frictionless horizontal surface. Because a frictionless surface does not offer, normal reaction.
- What are the conditions on which the coefficient of friction between two surfaces depends?
The coefficient of friction depends upon the nature of both surfaces in contact, its evenness and the surface temperature.