Newton's 2nd Law Problem, Angle exerted on block by surface

[Icycub]

Homework Statement

Problem:
A 12.0-kg block is pushed to the left across a rough horizontal surface by a force that is angled 30.0◦ below the horizontal. The magnitude of the force is 75.0 N and the acceleration of the block as it is pushed is 3.20 m/s^2 . What angle does the force exerted on the block by the surface make with the horizontal?

For this I have :
m= 12kg
theta= 30 degrees
applied force: 75 N
acceleration= 3.20 m/s^2

Homework Equations

For this problem I didn't use any equations, I only drew a picture and free body diagram (attached).

3. The Attempt at a Solution

Sorry if the image is big

I guess the main reason I didn't use any equations was because I am a little confused on what the question is asking for. Would the angle in question simply be 90 degrees? Isn't the force exerted on the block by the surface the normal force, which would be perpendicular to the horizontal?
Thank you for all the help.

 

[collinsmark]

Hello @Icycub,

Welcome to PF!

Homework Statement

Problem:
A 12.0-kg block is pushed to the left across a rough horizontal surface by a force that is angled 30.0◦ below the horizontal. The magnitude of the force is 75.0 N and the acceleration of the block as it is pushed is 3.20 m/s^2 . What angle does the force exerted on the block by the surface make with the horizontal?

For this I have :
m= 12kg
theta= 30 degrees
applied force: 75 N
acceleration= 3.20 m/s^2

Homework Equations

For this problem I didn't use any equations, I only drew a picture and free body diagram (attached).

3. The Attempt at a Solution
View attachment 112968 Sorry if the image is big

I guess the main reason I didn't use any equations was because I am a little confused on what the question is asking for. Would the angle in question simply be 90 degrees? Isn't the force exerted on the block by the surface the normal force, which would be perpendicular to the horizontal?
Thank you for all the help.

I think the problem statement is asking you to combine the normal force and the frictional force into a single "surface" force [Edit: it's a vector, don't forget]. Then, calculate the angle of this surface force with respect to the horizontal.

 

[Naso]

Yes, you are supposed to consider the reaction force to be the vectorial sum of the normal and frictional force (which some books call "tangent reaction")

 

[gneill]

Hi Naso,

could you send me the solution value? Just to check my own work :)

Sorry, that's not how things work here. If you'd like help checking your work you should start your own thread and post what you've done. Helpers will be happy to give you a hand once you've shown your work.

 

[Naso]

Hi Naso,

Sorry, that's not how things work here. If you'd like help checking your work you should start your own thread and post what you've done. Helpers will be happy to give you a hand once you've shown your work.

Oh alright, Edited!

 

[Icycub]

Alright thanks for the help, it makes sense that I have to combine the friction and normal force. I've worked out the problem, but I'm not sure if my answer is correct. In order to find the normal and kinetic force and then add them together, I set up two equations:

(note: I reversed the positive and negative signs for the x axis) --> left=positive right=negative

Σfx=ma
0-fk + facos(theta)=ma
fk=-ma +facos(theta)
fk= -(12)(3.2)+75cos(30)= 26.6 N

Σfy=0
N-fg-fasin(theta)=0
N=fg+fasin(theta)
N= (12)(9.8)+75sin(30)= 155 N

I'm confused because kinetic friction is positive. With my chosen coordinate system, I have it pointing in the negative x direction. Is this the correct value for kinetic friction?

 

[Naso]

That is the correct value; it is positive because you already considered it as being subtractive in the ∑Fx equation

 

[Icycub]

Alright thank you!