Question

To understand kinetic and static friction. A block of mass m lies on a horizontal table. The coefficient of static friction between the block and the table is μs. The coefficient of kinetic friction is μk, with μk<μs.a- Suppose you want to move the block, but you want to push it with the least force possible to get it moving. With what force F must you be pushing the block just before the block begins to move?Express the magnitude of F in terms of some or all the variables μs, μk, and m, as well as the acceleration due to gravity g.b- Suppose you push horizontally with half the force needed to just make the block move. What is the magnitude of the friction force?Express your answer in terms of some or all of the variables μs, μk, and m, as well as the acceleration due to gravity g.c- Suppose you push horizontally with precisely enough force to make the block start to move, and you continue to apply the same amount of force even after it starts moving. Find the acceleration a of the block after it begins to move.Express your answer in terms of some or all of the variables μs, μk, and m, as well as the acceleration due to gravity g.

Answer 1

Answer:

a) F = $\mu_smg$

b) F' = $\frac{F}{2}$

       = $\frac{\mu_smg}{2}$

       = 0.5 $\mu_smg$

c) $F_{applied} = \mu_smg$

so, $\mu_smg - \mu_kmg = ma$

  ⇒ a = $(\mu_s - \mu_k)$g

      a = ($\mu_s - \mu_k$)g

Explanation:

Friction acts on a body, when block moves on a surface or when the block have a tendency of motion on a surface. However, neither block is moving or there is a tendency of motion in this case.

So Friction F = 0

Answer 2

Answer: a) Fmax= μs.m.g  b) Ff=(μs.m.g /2  c) a= (μs.m.g - μk.m.g) / m

Explanation:

a) The friction force, is one of the components of the contact force, and can adopt any value to counteract an applied force, so the object doesn't move, till a maximum value, beyond which, if he applied force is larger, the object will start to move.

This limit value, is given by the following expression:

Fmax =  μs. N = μs. m.g

(This is valid only if the contact surface is horizontal).

 

b) As explained above, if the applied force is smaller than the limit value, the friction force will adopt the same value as the applied force, but of opposite direction, so due to the Newton's 2nd Law, the object remains at rest.

In this case, if the push is horizontal, with a force equal to the half of the limit value, friction force will be exactly equal to half the maximum friction force, as follows:

Ff = (μs. m.g) / 2

c) As the applied force is equal to the static friction force, once in movement, the opposing friction force is equal to Ffk = μk. m.g (where μk ∠ μs).

So, if the applied force is larger than this friction force (that always oppose to the relative movement between both surfaces in contact each other), the object will suffer an acceleration, which value is obtained from the Newton's 2nd Law, as follows:

Ext F = m .a ⇒ a = (μs.m.g -μk. m.g ) / m