A block with mass = 5.0 rests on a frictionless table and is attached by a horizontal spring ( = 130 ) to a wall. A second block, of mass = 1.35 , rests on top of . The coefficient of static friction between the two blocks is 0.40.
What is the maximum possible amplitude of oscillation such that will not slip off ?
assuming the spring is horizonal.
you have to figure the max force.
forceonTOP block=max spring force=k*D
where D is max amplitude
Now this cannot be greater than the friction force, 1.35*g*mu
k*D=1.35 g mu
d=1.35g*mu/k
check my thinking.
I get 0.041meters. this answer does not seem plausible?
To find the maximum possible amplitude of oscillation such that the second block will not slip off, we need to consider the forces acting on both blocks.
Let's consider the forces on block 2 first. The only horizontal force acting on it is the force of static friction between block 1 and block 2. The maximum value of static friction is given by:
\( F_{\text{friction}} = \mu_s \cdot F_{\text{normal}} \)
where \( \mu_s \) is the coefficient of static friction and \( F_{\text{normal}} \) is the normal force.
The normal force on block 2 is the weight of block 2 minus the upward force due to the spring:
\( F_{\text{normal}} = m_{2} \cdot g - F_{\text{spring}} \)
where \( m_{2} \) is the mass of block 2, \( g \) is the acceleration due to gravity, and \( F_{\text{spring}} \) is the force exerted by the spring.
The force exerted by the spring is proportional to the displacement of block 2 from its equilibrium position. By Hooke's Law, the force exerted by the spring is given by:
\( F_{\text{spring}} = -k \cdot x \)
where \( k \) is the spring constant and \( x \) is the displacement of block 2.
Combining these equations, the maximum force of static friction is:
\( F_{\text{friction}} = \mu_s \cdot (m_{2} \cdot g - F_{\text{spring}}) \)
Now, for block 1 to remain in equilibrium, the net force on block 1 should be zero. The only horizontal force acting on block 1 is the force exerted by the spring, \( F_{\text{spring}} \). The maximum force exerted by the spring is:
\( F_{\text{spring}} = k \cdot x \)
Therefore, the maximum amplitude of oscillation such that block 2 will not slip off is the value of \( x \) for which the force of static friction on block 2 is equal to the force exerted by the spring on block 1. In other words,
\( \mu_s \cdot (m_{2} \cdot g - F_{\text{spring}}) = k \cdot x \)
Substituting the expressions for \( F_{\text{spring}} \) and \( F_{\text{friction}} \),
\( \mu_s \cdot (m_{2} \cdot g - k \cdot x) = k \cdot x \)
Simplifying,
\( x = \frac{\mu_s \cdot m_{2} \cdot g}{2k + \mu_s \cdot m_{2} \cdot g} \)
Plugging in the given values:
\( x = \frac{(0.40) \cdot (1.35 \text{ kg}) \cdot (9.8 \text{ m/s}^2)}{2 \cdot (130 \text{ N/m}) + (0.40) \cdot (1.35 \text{ kg}) \cdot (9.8 \text{ m/s}^2)} \)
Calculating the answer, we find:
\( x \approx 0.028 \) meters
Therefore, the maximum possible amplitude of oscillation such that block 2 will not slip off is approximately 0.028 meters.