Dynamics and F=ma, where a is 0

[FaithAlyRose]

Homework Statement

A 24kg box is placed on a rough slope. Determine the magnitude of the frictional force exerted by the slope on the box if the box is moving downward at a constant velocity of 9.0m/s. The slope incline is 40°

Homework Equations

F=ma

The Attempt at a Solution

F = 24 x 0
a = 0 because speed is constant.
Resultant force = Initial force - frictional force

 

[vivekrai]

What is that attempt? You seem to have learned things in way too hurry. First draw the FBD of the block. See what are the forces acting on it. How can it's acceleration be put to zero etc.,

 

[vela]

Start with a free-body diagram. I encourage you to explain what you're doing instead of trying to make us guess what you reasoning is.

 

[FaithAlyRose]

Start with a free-body diagram. I encourage you to explain what you're doing instead of trying to make us guess what you reasoning is.

Well, as stated in the title, I used F=ma, so the working I came up with was F=24x0, with 24 being the mass of the box and 0 for the acceleration since the speed is constant.

 

[FaithAlyRose]

What is that attempt? You seem to have learned things in way too hurry. First draw the FBD of the block. See what are the forces acting on it. How can it's acceleration be put to zero etc.,

Haha sorry, I'm still very new to Physics, as I'm only 13. Our teacher just started teaching us, but he's far from thorough so we're struggling...

For the FBD, I drew
1) weight acting vertically downwards
2) friction acting opposite the direction of motion
3) a force acting perpendicular to the incline of the slope
4) the direction of motion

Thanks for offering your help! :)

 

[vela]

The Attempt at a Solution

F = 24 x 0
a = 0 because speed is constant.
Resultant force = Initial force - frictional force

What do you mean by "initial force"?

Well, as stated in the title, I used F=ma, so the working I came up with was F=24x0, with 24 being the mass of the box and 0 for the acceleration since the speed is constant.

I figured that's what you probably did, but it's better to hear you actually say it. Students can quite often come up with the correct answer for the wrong reasons.

For the FBD, I drew
1) weight acting vertically downwards
2) friction acting opposite the direction of motion
3) a force acting perpendicular to the incline of the slope
4) the direction of motion

Choose a set of axes and resolve the forces into x and y components. Then set up two equations:
\begin{align*}
\sum F_x &= ma_x \\
\sum F_y &= ma_y
\end{align*} where, as you noted, you have $a_x = a_y = 0$ since the box moves at constant speed.