Solving Block on Incline Homework: F_w, F_n, F_f, & Theta

[Chandasouk]

Homework Statement

A block lies on a plane raised an angle theta from the horizontal. Three forces act upon the block: F_w_vec, the force of gravity; F_n_vec, the normal force; and F_f_vec, the force of friction. The coefficient of friction is large enough to prevent the block from sliding .

Part A
Consider coordinate system a, with the x-axis along the plane. Which forces lie along the axes?

Part B
Which forces lie along the axes of the coordinate system b, in which the y-axis is vertical?

Now you are going to ignore the general rule (actually, a strong suggestion) that you should pick the coordinate system with the most vectors, especially unknown ones, along the coordinate axes. You will find the normal force, F_n_vec, using vertical coordinate system b. In these coordinates you will find the magnitude F_n appearing in both the x and y equations, each multiplied by a trigonometric function.

Part C
Because the block is not moving, the sum of the y components of the forces acting on the block must be zero. Find an expression for the sum of the y components of the forces acting on the block, using coordinate system b.
Express your answer in terms of some or all of the variables F_n, F_f, F_w, and theta.

Part D
Because the block is not moving, the sum of the x components of the forces acting on the block must be zero. Find an expression for the sum of the x components of the forces acting on the block, using coordinate system b.
Express your answer in terms of some or all of the variables F_n, F_f, F_w, and theta.

Part E
To find the magnitude of the normal force, you must express F_n in terms of F_w since F_f is an unknown. Using the equations you found in the two previous parts, find an expression for F_n involving F_w and theta but not F_f.


For Part A, the answer was F_f and F_n, but I really don't understand how because of what the question asks me to do. Is it because in that coordinate plane, Normal Force and Frictional force are directly on the axes already there?

 

[tiny-tim]

Hi Chandasouk!

(try using the X₂ tag just above the Reply box )

Part A
Consider coordinate system a, with the x-axis along the plane. Which forces lie along the axes?

Part B
Which forces lie along the axes of the coordinate system b, in which the y-axis is vertical?
…
For Part A, the answer was F_f and F_n, but I really don't understand how because of what the question asks me to do. Is it because in that coordinate plane, Normal Force and Frictional force are directly on the axes already there?

Yes. It's not a trick question. It really is only asking which forces lie along each axis.

As the question goes on to say, usually you'd use the "best" axes, judging by exactly that criterion.

But the question has decided, in parts C to E, to give you plenty of practice at resolving forces into components, and at solving differential equations … which you wouldn't get if it let you do it the easy way!

What do you get? ​

 

[tomcue]

For Part B, the forces that lie along the axes are F_w and F_f. In the vertical coordinate system, the x-axis is perpendicular to the y-axis, so only the forces that are parallel to the y-axis (F_w and F_f) will have components along that axis.

For Part C, the sum of the y components of the forces must be zero because the block is not moving in the y-direction. This means that the upward force from the normal force must be equal to the downward force from the weight of the block, expressed as F_n*sin(theta) = F_w*cos(theta).

For Part D, the sum of the x components of the forces must be zero because the block is not moving in the x-direction. This means that the force of friction must be equal and opposite to the component of the weight of the block that is parallel to the plane, expressed as F_f = F_w*sin(theta).

For Part E, we can substitute the expression for F_f found in Part D into the expression for F_n found in Part C. This gives us F_n = F_w*cos(theta)/cos(theta), which simplifies to just F_n = F_w. So, the magnitude of the normal force is equal to the weight of the block.