A block of mass M resting on a 21.5° slope is shown. The block has coefficients of friction μs=0.605 and μk=0.344 with the surface. It is connected via a massless string over a massless, frictionless pulley to a hanging block of mass 2.57 kg. What is the minimum mass M1 that will stick and not slip?

well, tension is 2.57g up the slide

static friction is .605*M1*g*cosTheta, down the slide
gravity down the slide=M1*g*sinTheta

so,

2.57g=.605M1*g*cosTheta-M1*g*sinTheta

solve for M1

check my thinking and typing.

Bob was mostly right, however when he took into consideration of the gravity on the down side he forgot to input the static friction that goes against gravity pulling the block down the slope.

Tension= (Us*(mass*9.8cos(theta))-((mass*9.8sin(20))-(Us*(mass*9.8cos(20))

So without numbers and only variables:
T=(Us*N)-(g)-(Us*N)

obviously, when you have a slope, you have to factor the theta in, as I did in the first equation.

Well, first of all, let's just appreciate the fact that this block has quite the identity crisis. It's trying to decide whether it wants to stick or slip. I mean, I feel you, little block. I can barely decide which socks to wear in the morning.

Now, let's tackle this problem. To determine the minimum mass M1 that will stick and not slip, we need to consider the forces acting on the block.

We have the force of gravity pulling the hanging block downwards and creating tension in the string. We also have the normal force, the force of friction, and the force pulling the block down the slope.

When the block is at the verge of slipping, the force of friction reaches its maximum value, given by the equation Ffriction (max) = μs * N, where N is the normal force.

But here comes the fun part (and by fun, I mean mathematically challenging). We need to break down the force of gravity into components parallel and perpendicular to the slope.

The component of gravity parallel to the slope is m1 * g * sin(θ), where θ is the angle of the slope. The normal force is equal to the component of gravity perpendicular to the slope, which is m1 * g * cos(θ).

Since the block is at the verge of slipping, we know that the force of friction is equal to μs * N, which is equal to μs * m1 * g * cos(θ).

But that's not all, folks! We also have the tension in the string, which is equal to the weight of the hanging block, which is m2 * g.

Taking all of this into account, we can write the equation:

μs * m1 * g * cos(θ) = m2 * g + m1 * g * sin(θ)

Now, we just plug in the given values and solve for m1:

0.605 * m1 * 9.8 * cos(21.5°) = 2.57 * 9.8 + m1 * 9.8 * sin(21.5°)

And after some algebraic gymnastics, we find that m1 is approximately equal to 5.23 kg.

So, the minimum mass M1 that will stick and not slip is around 5.23 kg. Remember, little block, stick to your decisions! Don't be a slippery slope!

To find the minimum mass M1 that will stick and not slip, we need to consider the balance of forces acting on the block on the slope.

First, let's draw a free body diagram of the block on the slope:

/|
/ |
/ |
/ θ |
/____________|

^ Ff
|
| Fg
|
|_ Fy

The forces acting on the block are:
1. The force of gravity (Fg) acting vertically downward, which can be divided into two components:
- Fg_x: The component parallel to the incline.
- Fg_y: The component perpendicular to the incline.

2. The normal force (Fn) acting perpendicular to the incline.

3. The force of friction (Ff), which opposes the motion, acting parallel to the incline.

Now, let's analyze the forces acting on the system. Since the block is not slipping, the force of friction must be equal to or greater than the force required to initiate slipping. Therefore, the force of friction can be calculated as:

Ff = μs * Fn,

where μs is the coefficient of static friction.

The normal force (Fn) can be calculated as Fn = Fg_y = M * g * cos(θ), where θ is the angle of the slope.

The force of gravity acting parallel to the incline can be calculated as Fg_x = M * g * sin(θ).

Next, we need to consider the tension in the string. The hanging block exerts a force (T) on the system, which can be calculated as T = M1 * g.

Now, let's equate the forces and solve for M1:

Ff = T

⇒ μs * Fn = T

⇒ μs * (M * g * cos(θ)) = M1 * g

⇒ M1 = (μs * M * cos(θ)) / (1 - μs * sin(θ))

Plugging in the given values:
μs = 0.605,
θ = 21.5°,
M = 2.57 kg,
g ≈ 9.8 m/s², (acceleration due to gravity)

M1 = (0.605 * 2.57 * cos(21.5°)) / (1 - 0.605 * sin(21.5°))

Calculating this expression will give you the minimum mass M1 that will stick and not slip.