Roy pushed a table with a force of 5 N. The magnitude limiting value of static frictional force is 15 N. The minimum additional force required to move the table is ______________N.
Answers
Answer:
On the point of sliding and when sliding occurs:
F = μR
where F is the friction force acting between two surfaces in contact,
μ is the coefficient of friction, and R is the normal contact force.
(Note that sometimes N is used instead of R for the normal contact force.)
You can use the friction model to determine whether a body will slide, what
force will be required to make a body slide, or to find the coefficient of
friction between two surfaces in contact.
Example
A 5 kg box on a horizontal table is pushed by a horizontal force
of 15 N as shown on the right.
If the coefficient of friction is 0.4, will the box move?
Think about
What is the smallest force that could make the box slide along the table?
Solution
The sketch shows the forces acting on the box.
Note that the weight of a box of mass 5 kg is 5g where g = 9.8 ms–2
Since the vertical forces are in equilibrium, R = 5g
Therefore the maximum possible friction is F = μR = 0.4 5g = 19.6 N
The pushing force, 15 N, is less than this and so cannot overcome the
friction.
The box will not move.