Understanding Free Body Diagrams: How to Determine Angle Placement?

[smg528]

So I'm a little confused about free body diagrams. I understand the concept, but when it comes to drawing, how do you know where the angle goes? I don't know if this is obvious, but physics is not really my cup of tea... Thanks in advance.

 

[rl.bhat]

Attach the axis to the body so that the x-axis is in the direction of the motion of the body.

 

[smg528]

But how come sometimes, the angle is above the x-axis with the normal force and other times it's drawn below with gravity even though the picture given looks the same in different problems?

 

[rl.bhat]

You have to draw the free body diagram for individual body. Identify all forces acting on the body. Depending on the direction of the motion, each one will be different form each other. If you can post a problem, I can help you.

 

[smg528]

Ok, here's two problems with similar pictures, but in the first free body diagram, the angle is below the x-axis and in the second, the angle is above the x-axis. I'm really confused about that.

1. A skier of mass 65 kg glides down a slope at angle = 32. Find the skier's acceleration and the force the snow exerts on the skier. The snow is so slipper you can neglect friction.

2. A starting gate acts horizontally to restrain a 60-kg ski racer on a frictionless 30-degree slope. What horizontal force does the starting gate apply to the skier?

 

[rl.bhat]

1)Skier move in the downward direction So x-axis is parallel to slope.
What are the forces acting on the skier?
i) mg*sinθ along the slope.
ii) mg*cosθ perpendicular to the slope.
iii) Normal force due to the slope.
Net force is along the slope.
2) It is similar to the first one. Only addition is the horizontal gate.
mg*sinθ is the force exerted by the skier on the gate along the slope.
Now attach the axis to the gate. x-axis horizontal . Find the horizontal component of mg*sinθ. That will the reaction of the gate on the skier.

 

[smg528]

But for the second problem, it says that the x component is -horizontal F + nsin(theta)=0 and the y component is ncos(theta) - mg=0

 

[rl.bhat]

But for the second problem, it says that the x component is -horizontal F + nsin(theta)=0 and the y component is ncos(theta) - mg=0

Here what is n? What is its expression?

 

[smg528]

n is the normal force

 

[rl.bhat]

In the second problem, what is the orientation of x an y-axis?
You say that y-component is n*cosθ - mg = 0. That means n > mg. How is that? Normal force must appear in pair. Which one is action and which one is reaction?