I understand this, but what I don't understand is why we cant solve for the normal force using mgcos(15). On the FBD cos is the y component which should give us the N force. When I do 490.5cos(15) I get 424.79N which is different using the sys-sold method (which you got 494.95N). Any idea why this method does not get me the correct normal force, thank you!
Good question. Since we are doing the analysis for the finding the Force needed to keep the box from sliding down, we can conclude the direction of motion is "sliding down". So the friction force would point upward.
Would it be? I think I made it a habit to always translate the axes along the incline for x and perpendicular to the incline for y. I may have overdone it for this question.
Yes not always. In this case, the normal force would have to counter act the weight component in the y’ direction (wcos(thetha)) and the applied force component along y’ (Fy’). So the normal force N = wcos(thetha) + Fy’
I underestimated the question earlier🙂, I solved it without considering the effect of the 'F' on the reaction force. Thank you for this. Also, we could have resolved the forces onto the regular X and Y axis.
Hello Ahmad, thank you for attempting the question. When solving for the force required to move the box up the incline, I got a value F = 474.27 N and N = 661.93 N. I am not 100% sure this is correct though.
I understand this, but what I don't understand is why we cant solve for the normal force using mgcos(15). On the FBD cos is the y component which should give us the N force. When I do 490.5cos(15) I get 424.79N which is different using the sys-sold method (which you got 494.95N). Any idea why this method does not get me the correct normal force, thank you!
What an awesome video!
Why is the direction of frictional force same as applied force? Isn't it supposed to be opposite to the applied force, if so answer will be 474N
Good question.
Since we are doing the analysis for the finding the Force needed to keep the box from sliding down, we can conclude the direction of motion is "sliding down". So the friction force would point upward.
Is it not simpler to use normal X&Y axises in this question?
I was thinking the same
Would it be? I think I made it a habit to always translate the axes along the incline for x and perpendicular to the incline for y. I may have overdone it for this question.
Why not just calculate the normal force?
Does this mean normal force is not always W cos theta?
Yes not always. In this case, the normal force would have to counter act the weight component in the y’ direction (wcos(thetha)) and the applied force component along y’ (Fy’). So the normal force N = wcos(thetha) + Fy’
I underestimated the question earlier🙂, I solved it without considering the effect of the 'F' on the reaction force. Thank you for this. Also, we could have resolved the forces onto the regular X and Y axis.
15322 N
Hello Ahmad, thank you for attempting the question. When solving for the force required to move the box up the incline, I got a value F = 474.27 N and N = 661.93 N. I am not 100% sure this is correct though.
@@directhubfeexam I got the same answer, but Farooq mentioned that the friction force is upward. In that case, the answer will be different.