A large block P executes horizontal simple harmonic motion as it slides across a frictionless surface with a frequency f = 1.7 Hz. Block B rests on it, as shown in the figure, and the coefficient of static friction between the two is ìs=0.64. (b) What is the maximum force of static friction for a 6.7 kg mass?
To find the maximum force of static friction, we need to consider the criteria for static friction to be at its maximum. In this case, the maximum force of static friction is the force at which the block B is on the verge of sliding off block P.
First, let's find the maximum acceleration (amax) that can be supported by the harmonic motion of block P. We know that the frequency of the simple harmonic motion is f = 1.7 Hz.
The frequency of simple harmonic motion (f) is related to the maximum acceleration (amax) by the formula:
f = (1/2π) * √(k/m)
where k is the spring constant and m is the mass of the block.
Rearranging the formula, we can solve for amax:
amax = (4π²f²) * m
Next, let's find the maximum force of static friction (fstatic_max) using the equation:
fstatic_max = ìs * m * g
where ìs is the coefficient of static friction, m is the mass, and g is the acceleration due to gravity.
Now we can calculate the maximum force of static friction:
fstatic_max = ìs * m * g
Let's substitute the given values into the formula:
ìs = 0.64
m = 6.7 kg
g = 9.8 m/s²
fstatic_max = 0.64 * 6.7 kg * 9.8 m/s²
Now, calculate the value:
fstatic_max = 42.7776 N
Therefore, the maximum force of static friction for a 6.7 kg mass is approximately 42.78 N.