A block of mass M1 = 160.0 kg sits on an inclined plane and is connected to a bucket with a massless string over a massless and frictionless pulley. The coefficient of static friction between the block and the plane is ms = 0.52, and the angle between the plane and the horizontal is q = 47°. Mass M2 (mass of the bucket) can be changed by adding or taking away sand from the bucket. What is the maximum value of M2 for which the system will remain at rest?
Can someone please give me a little help?
First, break the forces on M1 into normal and down the plane forces.
Down the plane is mgSinTheta, and friction down the plane of -mg*mu*CosTheta.
A system equations: To stop the bucket from going up, the weight of the bucket must equal the forces down the plane.
M2*g=mgSinTheta-mg*mu*CosTheta
check my thinking.
That's right.
Now it asks for the minimum value of M2 for which the system will remain at rest. Do I still use mgSinTheta-mg*mu*CosTheta?
In the first case, maximum M2, both the component of weight down the plane and friction are in the down plane direction. They add, no minus sign.
If the thing is to move the other way, down the ramp, the friction is opposite in direction, up the ramp. So you change the sign of that second term, the friction one. NOW you have a minus sign
Ok I got it, thanks.
The last part is this: Suppose the coefficient of kinetic friction between the block and the inclined plane is muk = 0.36. If M2 = 217.1 kg, what is the magnitude of the acceleration of M1?
Which equation am I using for this?
To find the maximum value of M2 for which the system will remain at rest, we need to consider the forces acting on the block on the inclined plane.
Let's break down the forces on the block:
1. The weight of the block (M1 * g), where g is the acceleration due to gravity.
2. The normal force (N) exerted by the inclined plane on the block perpendicular to the surface.
3. The force of static friction (Fs) between the block and the inclined plane.
We need to find the maximum static friction force that can be exerted by the surface of the incline to keep the block at rest. This occurs when the force of static friction reaches its maximum value, which is directly proportional to the normal force: Fs_max = μs * N.
Here, μs is the coefficient of static friction given as 0.52.
To start solving the problem, let's resolve the weight (M1 * g) into two components:
1. The component parallel to the inclined plane, which is M1 * g * sin(q).
2. The component perpendicular to the inclined plane, which is M1 * g * cos(q).
Since the system is at rest, the force of static friction acting up the incline must exactly balance the component of the weight acting down the incline (M1 * g * sin(q)). Therefore, we can set up the following equation:
μs * N = M1 * g * sin(q)
We also know that N, the normal force, is equal to the perpendicular component of the weight (M1 * g * cos(q)).
So, we can substitute N in the above equation:
μs * (M1 * g * cos(q)) = M1 * g * sin(q)
Now, we can solve for M2, the mass of the bucket:
M1 * g * cos(q) * μs = M2 * g
We can cancel out g from both sides of the equation:
M1 * cos(q) * μs = M2
Finally, substituting the given values, we can calculate the maximum value of M2.