A wooden block with mass 0.26 kg rests on a horizontal table, connected to a string that hangs vertically over a friction-less pulley on the table's edge. From the other end of the string hangs a 0.12 kg mass.
A) What minimum coefficient of static friction ? s between the block and table will keep the system at rest?
B) Find the block's acceleration if ? k =0.18.
To find the minimum coefficient of static friction (µs) between the block and the table, we need to consider the forces acting on the block.
A) The forces acting on the block include its weight (mg) and the tension in the string (T). The block is at rest, so the net force acting on it is zero.
First, let's analyze the forces on the hanging mass (0.12 kg):
- The tension in the string (T) acts upward.
- The weight of the hanging mass (mgh) acts downward.
Since the hanging mass is in equilibrium, the tension (T) is equal to the weight of the hanging mass:
T = mgh
Now, let's analyze the forces on the wooden block (0.26 kg):
- The weight of the block (mg) acts downward.
- The tension in the string (T) acts upward.
- The frictional force (f) acts horizontally, opposing the impending motion of the block.
Since the block is at rest, the net force acting on it is zero. Therefore, the sum of the horizontal forces (friction and tension) must be equal to the sum of the vertical forces (weight):
µs * N = mg [Equation 1]
N is the normal force, which is equal to the weight of the block because the block is on a horizontal table:
N = mg
Substituting N = mg into Equation 1, we get:
µs * mg = mg
Simplifying the equation:
µs = 1
Therefore, the minimum coefficient of static friction (µs) required to keep the system at rest is 1.
B) To find the block's acceleration (a) when the kinetic friction coefficient (µk) is given, we also need to consider the forces acting on the system.
The weight of the hanging mass (mgh) creates a tension (T) in the string, which acts as an opposing force to the motion of the block. The net force on the block is the difference between the tension (T) and the frictional force (f), both acting in the same direction:
Net Force = T - f
The tension in the string (T) is again given by:
T = mgh
The frictional force (f) can be calculated using the equation:
f = µk * N, where N is the normal force.
The normal force (N) is still equal to the weight of the block because it is on a horizontal table:
N = mg
Substituting N = mg into the equation for frictional force:
f = µk * mg
Now, substituting T = mgh and f = µk * mg into the equation for net force:
Net Force = mgh - µk * mg
Using Newton's second law (F = ma) and rearranging the equation:
ma = mgh - µk * mg
Simplifying:
a = g(h - µk)
Therefore, the block's acceleration (a) when the kinetic friction coefficient (µk) is given is a = g(h - µk).