Will Box Y Remain in Equilibrium?

[aerograce]

Homework Statement

A 8-kg homogeneous box X is resting against another 20-kg homogeneous box Y at
an angle such that its diagonal BC is horizontal as shown in Figure Q1. The angle
of static friction between box X and the floor is 45° and that between box Y and the
floor is 30°. Treat the contact surface between the two boxes as smooth. For
practical simplicity, use g = 10 m/s2.
Figure: View the attachment

(a) Determine whether box X is in equilibrium.
(3 marks)
(b) Determine whether box Y is in equilibrium with respect to sliding along the floor
and rotation about the corner A.
(4 marks)

Homework Equations

\mu=\frac{F}{N};
Tan\alpha=\mu

The Attempt at a Solution

Solving for Question b:
Identify that Box X is a three force body, and if it has to reach equilibrium state, the three forces have to pass through the center of mass. Draw a reaction force R(A) from the bottom of box X to its center of mass. Based on calculation, the angle formed by the R(A) with respect to the ground is 48.2°, which is more than the angle of static friction, and hence Box X is not equilibrium.

My problem is that, when I went to analyze Box Y, I cannot get sufficient values to analyse it as Box X is not at equilibrium and hence the normal contact force between them cannot be calculated.

Any thoughts? Thank you so much!

 

[tiny-tim]

hi aerograce!

Based on calculation, the angle formed by the R(A) with respect to the ground is 48.2°, which is more than the angle of static friction, and hence Box X is not equilibrium.

isn't it the other way round?

if the angle from the ground was 90°, that would be ok wouldn't it?

 

[aerograce]

hi aerograce!


isn't it the other way round?

if the angle from the ground was 90°, that would be ok wouldn't it?

Hello! Could you attempt this problem? I recalculate my forces and found box X still not in equilibrium. The required minimum angle of friction is 53.2 degrees, which is more than angle of the static friction provided. Hence box X will not be in equilibrium.

My friend looks at this problem another way:
He takes moments of Box X about the point where it touches the ground, which gives him the reaction forces between two boxes. And this reaction force is 71N, which is less than the maximum static friction here(80N), hence it is in equilibrium. And my question is, if reaction force is 71N, the static friction force will be 71N as well. And the three forces will definitely not converge at one single point(according to geometry).

Sorry I am very confused now and I am looking for your help:) Thanks a million!

 

[tiny-tim]

hello aerograce!

My friend looks at this problem another way:
He takes moments of Box X about the point where it touches the ground, which gives him the reaction forces between two boxes. And this reaction force is 71N, which is less than the maximum static friction here(80N), hence it is in equilibrium. And my question is, if reaction force is 71N, the static friction force will be 71N as well. And the three forces will definitely not converge at one single point(according to geometry).

yes they do converge

the three forces are the horizontal reaction force at B, the weight, and the reaction force from the ground

the first two forces obviously go through the centre of mass

the third force (which is mg down and 71 N horizontally) also goes through the centre of mass