1 A block of mass mb sits atop a at wooden plank of mass mp on a plane inclined an angle theta with respect to thehorizontal. The two are supported near the bottom of the plane by a spring of spring constant k that is in contactwith the plank but not the block. If the coecient of friction between the plank and the plane is µ1 and the coefficent of friction between the block and the plank is µ2 answer the following questions.
a. Find an expression for the maximum compression of the spring such that the plank and block remain at rest.
b. If the spring is compressed three times the amount you found in part (a) find the initial acceleration of theplank and the acceleration of the block given the following values: k = 4025N/m, mp = 1.2kg, mb = 4.3kg, µ1 = 0.12, µ2 = 0:07.
a. To find the maximum compression of the spring such that the plank and block remain at rest, we need to calculate the net force acting on the plank and block system in both the horizontal and vertical directions and ensure that they sum up to zero.
Let's start by analyzing the forces in the vertical direction. The weight of the plank and block system can be calculated as:
Weight of the plank (Fp) = mp * g (g is the acceleration due to gravity)
Weight of the block (Fb) = mb * g
The normal force acting on the plank and block system is equal to the component of the weight perpendicular to the inclined plane, which can be calculated as:
Normal force (Fn) = (mp + mb) * g * cos(theta)
The friction force between the plank and the inclined plane can be calculated as:
Friction force (Ff1) = µ1 * (mp + mb) * g * sin(theta)
Now, let's analyze the forces in the horizontal direction. The force exerted by the spring can be calculated as:
Spring force (Fs) = k * x (x is the compression of the spring)
The friction force between the block and the plank can be calculated as:
Friction force (Ff2) = µ2 * (mp + mb) * g * cos(theta) (since the block slides on the plank)
Since the system is at rest, the net force in both the horizontal and vertical directions must be zero. Therefore, we can write the following equations:
Horizontal direction: Fs - Ff2 = 0
Vertical direction: (mp + mb) * g - Fn - Ff1 = 0
Substituting the expressions for each force, we get:
k * x - µ2 * (mp + mb) * g * cos(theta) = 0
(mp + mb) * g - (mp + mb) * g * cos(theta) - µ1 * (mp + mb) * g * sin(theta) = 0
Now, solve these equations to find the maximum compression of the spring (x) such that the plank and block remain at rest.
b. If the spring is compressed three times the amount found in part (a) (3x), we need to calculate the initial acceleration of the plank and the acceleration of the block.
The net force acting on the system in the horizontal direction can be calculated as:
Net force (Fnet) = Fs - Ff2
Fnet = k * (3x) - µ2 * (mp + mb) * g * cos(theta)
The initial acceleration of the system is given by Newton's second law:
Fnet = (mp + mb) * a_initial
So, we can write:
k * (3x) - µ2 * (mp + mb) * g * cos(theta) = (mp + mb) * a_initial
The acceleration of the block can be calculated using the friction force and the weight of the block:
Acceleration of the block = (µ2 * (mp + mb) * g * cos(theta)) / mb
And that's how you can find the initial acceleration of the plank and the acceleration of the block given the provided values.
To solve this problem, we need to analyze the forces acting on the block and the plank and apply the principles of static equilibrium and Newton's second law. Let's break it down step by step:
a. For the system to remain at rest, the net force and net torque acting on it must be zero.
1. Forces along the incline:
The force component acting parallel to the incline is the weight of the block mb * g * sin(theta). This force is opposed by the friction force between the plank and the plane, which is µ1 * (mb + mp) * g * cos(theta), where g is the acceleration due to gravity.
2. Forces perpendicular to the incline:
The normal force N acting on the plank is the sum of the vertical components of the weight of the block and the plank, which is (mb + mp) * g * cos(theta). This normal force is balanced by the vertical component of the spring force, k * x (where x is the compression of the spring), and the weight component acting perpendicular to the incline, (mb + mp) * g * sin(theta).
Equating forces along the incline:
mb * g * sin(theta) = µ1 * (mb + mp) * g * cos(theta)
Equating forces perpendicular to the incline:
(mb + mp) * g * cos(theta) = k * x + (mb + mp) * g * sin(theta)
Solving these two equations simultaneously will give an expression for the maximum compression of the spring, x.
b. To find the initial acceleration of the plank and the block when the spring is compressed three times the amount found in part (a), we need to consider the forces acting on the system.
1. Forces along the incline:
The force component acting parallel to the incline is still the weight of the block mb * g * sin(theta). This force is opposed by the friction force between the plank and the plane, which is µ1 * (mb + mp) * g * cos(theta).
2. Forces perpendicular to the incline:
The normal force N acting on the plank is still (mb + mp) * g * cos(theta), balanced by the vertical component of the spring force, k * x (where x is the compression of the spring), and the weight component acting perpendicular to the incline, (mb + mp) * g * sin(theta).
In addition to these forces, there is now a force opposing the motion of the block due to the compression of the spring. This force can be calculated as -k * (3 * x) since the spring is compressed three times the amount found in part (a).
Applying Newton's second law in the direction along the incline:
mb * g * sin(theta) - µ1 * (mb + mp) * g * cos(theta) = (mb + mp) * a
where a is the acceleration of the system.
Applying Newton's second law perpendicular to the incline:
(mb + mp) * g * cos(theta) = k * x + (mb + mp) * g * sin(theta)
Solving these two equations simultaneously will give the initial acceleration of the plank (a) and the acceleration of the block. Substitute the given values of k, mp, mb, µ1, and µ2 to find the final numerical values.