Pulley problem with incline plane

[putongren]

Consider two blocks connected by an ideal rope that passes through an ideal pulley fixed on the corner of the wedge as shown in the figure. The blocks have masses m1 and m2. The block of mass m1 is on the incline surface (which is frictionless) while the block of mass m2 hangs up in the air. Find the angle θ at which the two blocks are in equilibrium (i.e., not moving)..

1. Homework Statement
N = Normal =m₁g
T = m₂g

Homework Equations

f = ma

The Attempt at a Solution

x - direction[/B]
N sin θ = T cos θ
m₁g sin θ = m₂g cos θ
cos θ = (m₁/m₂)sin θ

y - direction

N cos θ + T sin θ = m₁g
m₁g cos θ + m₂g sin θ = m₁g

substitute cos θ = (m₁/m₂)sin θ

(m²₁ / m₂) sin θ + m₂ sin θ = m₁

sin θ = m²₁ m₂ / m₁g + m₂

θ = sin^-1(m²₁ m₂)/ m₁g + m₂


But the angle is actually something else according to the answer key.

 

[Simon Bridge]

N = Normal =m1g

... check this: N us usually the force acting normal to a surface - ##m_1g## is gravity on ##m_1## which is not normal to any of the surfaces in the diagram.

In your working, you should specify how you are using your axes... where are you placing the x and y axis?
I think you have got the directions of the forces mixed up - just draw the forces in as arrows, and use trigonometry to work out the right equations.
i.e.
"N sin θ = T cos θ"
... suggests that you are using x and y as horizontal and vertical ... you'll find the maths easier if the x-axis for m1 points upwards along the ramp.

 

[Les talons]

(m21 / m2) sin θ + m2 sin θ = m1

sin θ = m21 / m2 / m1g + m2

How did you get from the above line to the bottom line? What is the order of operations in the bottom line? What order of divisions do you want to have happen?

[tex] \frac{m_{1}{}^{2}}{m_2} \sin \theta + m_2 \sin \theta = m_1 [/tex]

Try to simplify and then factor out the sine term.

 

[putongren]

How did you get from the above line to the bottom line? What is the order of operations in the bottom line? What order of divisions do you want to have happen?

[tex] \frac{m_{1}{}^{2}}{m_2} \sin \theta + m_2 \sin \theta = m_1 [/tex]

Try to simplify and then factor out the sine term.

Hey thanks for the feedback ... I defined the inclined ramp as the axis.[tex] m_1g \sin \theta = m_2 g [/tex]

[tex] \sin \theta = m_2 / m_1 [/tex]

[tex] \theta = \arcsin{m_2 / m_1} [/tex]

Which is the answer.

 

[Simon Bridge]

Hey thanks for the feedback ... I defined the inclined ramp as the axis.

If the x-axis is along the ramp for m1, then the normal force N has no component along the x-axis and the tension T has no component along the y axis.
So whichever way you look at it: "N sin θ = T cos θ" cannot be correct.

For m1, the tension T points upwards and to the left (the +x direction) ... the normal force N points upwards along the y-axis ... and the weight (gravity) W points directly downwards (this is the only force with components in x and y directions). Since the block is motionless, ##\vec N + \vec W + \vec T = 0##
That should make a difference to your equations.