# How Does the Angle of Inclination Affect Motion in Rigid Body Dynamics?

• fedecolo
In summary: But is there a unique value of α that permits it to be non-zero?In summary, the problem is a statics problem with a block of mass M, a pulley of mass M and radius R, and a second block of mass 2M on an inclined plane with angle α and static friction μ. The wire does not slip on the pulley and there is friction between the wire and the pulley. The objective is to find the condition that α must satisfy for the block 1 to move downward. The pulley is accelerating and the difference in tensions between the wire and the pulley must be taken into account. A helpful approach is to express the rotational acceleration of the pulley in terms of the linear acceleration of
fedecolo

## Homework Statement

In the figure there are a block 1 of mass ##M##, a pulley (disk) of mass ##M## and radius ##R## and a second block of mass ##2M## on an inclined plane with an angle ##\alpha## with respect to the horizontal. On this plane the static friction is ##\mu##.
The wire does NOT slip on the pulley (so there is friction on the pulley too, but it says that the tension ##T_1 \not= T_2##).

What is the condition that ##\alpha## must satisfy such that the block 1 moves downward?

## The Attempt at a Solution

I wrote the diagram of forces for all the elements (with the systems of reference in the figure) and the torques with respect to the center of mass of the pulley.

For the block 1: ##T_1 - Mg= M(-a)##

For the pulley:
$$\begin{cases} R_y -Mg -T_1 -T_2 \sin\alpha=0 \\ -R_x +T_2 \cos\alpha=0 \\ \end{cases}$$

##R## is the vincolar reaction of the pulley support.

For the block 2 (the system of reference is different!)
$$\begin{cases} T_2-f_s -2Mg\sin\alpha =2Ma \\ 2Mg\cos\alpha -N=0 \end{cases}$$

Now using the fact that ##\sum \vec{M} = I_{CoM} \cdot \alpha_z## (##\alpha_z=\frac{a}{R}## is the angular acceleration of the pulley, sorry for the bad notation similar to the angle) I find the torques with respect to the center of mass of the pulley:
$$\hat{u_z}: RT_1 -RT_2 \sin\alpha = 1/2 MR^2 \alpha_z$$

Now I can't solve this big system!
The condition is ##T_2 -2Mg\sin\alpha +2Ma \geq f_{s,MAX} = 2Mg\cos\alpha \cdot \mu##.

It seems that I must use the ##R## to find ##T_1## or ##T_2## but I have two different ##R_x## and ##R_y##...

Any help?

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fedecolo said:
What is the condition that α must satisfy such that the block 1 moves downward?
Are there more parts to the question? If it is only that then this is a statics problem, no? So at the critical value of α the tensions will be equal unless there is axial friction in the pulley, and we have no info on that.

haruspex said:
Are there more parts to the question? If it is only that then this is a statics problem, no? So at the critical value of α the tensions will be equal unless there is axial friction in the pulley, and we have no info on that.
Yes, it's statics problem.
The other question asks what are the accelerations of the blocks if the condition is satisfied.

The pulley can rotate without friction around its axis, but there is friction between the wire and the pulley, so the 2 tensions are different.

fedecolo said:
there is friction between the wire and the pulley, so the 2 tensions are different.
I have not seen a pulley problem yet that permits any slipping between rope and pulley. There is always friction there. Consider the torque balance about the pulley's axle. If the pulley is not accelerating, what does that tell you about the tensions?

haruspex said:
I have not seen a pulley problem yet that permits any slipping between rope and pulley. There is always friction there. Consider the torque balance about the pulley's axle. If the pulley is not accelerating, what does that tell you about the tensions?
But the pulley is accelerating, because the block 1 must move downward!

fedecolo said:
But the pulley is accelerating, because the block 1 must move downward!
If the pulley is accelerating, that tells you something about the difference in the tensions.

A useful starting approach would be to express the rotational acceleration of the pulley in terms of the linear acceleration of the block. The objective would be to then use this to find a formula for the tension difference in terms of the block acceleration.

It seems that you almost have that:
$$\hat{u_z}: RT_1 -RT_2 \sin\alpha = 1/2 MR^2 \alpha_z$$
However, that does not look correct to me. What is the ##\sin \alpha## doing there?

fedecolo said:
But the pulley is accelerating, because the block 1 must move downward!
It is asking for the boundary condition, i.e. the circumstance in which it barely moves downwards. That means its acceleration is arbitrarily small, so the difference in the tensions will be arbitrarily small.

jbriggs444

## 1. What is a rigid body in physics?

A rigid body in physics is an idealized concept of an object that does not deform or change shape when subjected to external forces. It is assumed to be made up of particles that maintain a fixed distance and orientation relative to each other.

## 2. What is the difference between a rigid body and a non-rigid body?

A non-rigid body is an object that can deform or change shape when subjected to external forces, while a rigid body maintains its shape and size. In real-world situations, most objects exhibit some degree of flexibility, making them non-rigid bodies.

## 3. What is a rigid body dynamics problem?

A rigid body dynamics problem is a type of physics problem that involves the study of the motion and forces acting on a rigid body. It requires the application of principles such as Newton's laws of motion and conservation of energy and momentum to determine the motion of the body under various conditions.

## 4. What are some real-world applications of rigid body dynamics?

Rigid body dynamics has various applications in fields such as engineering, robotics, and biomechanics. For example, it is used to study the motion and stability of structures, the design and control of robots, and the movement of human joints and muscles.

## 5. What are the main challenges in solving rigid body dynamics problems?

The main challenges in solving rigid body dynamics problems include accurately modeling the forces and torques acting on the body, considering the effects of external factors such as friction and air resistance, and dealing with complex or nonlinear systems. Additionally, finding the initial conditions and solving the equations of motion can be mathematically challenging.

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