MHB Force on 10 Kg Block on 51° Inclined Plane

[JessiMen]

A 10 Kg block lies on a smooth plane inclined at 51 degrees. What force parallel to the incline would prevent the block from slipping?

 

[MarkFL]

Can you post your progress so far, so our helpers can see how best to help?

 

[JessiMen]

oh I skipped this question in my review because i didnt know how to do it there is also a question b to it:

b) An object is dragged 6 m up a ramp under a constant force of 27 N applied at an angle of 32 degrees to the ramp. Calculate the work done

 

[topsquark]

oh I skipped this question in my review because i didnt know how to do it there is also a question b to it:

b) An object is dragged 6 m up a ramp under a constant force of 27 N applied at an angle of 32 degrees to the ramp. Calculate the work done

...And what have you been able to do so far?

-Dan

 

[JessiMen]

for this question nothing because i have no idea how to start it.

 

[MarkFL]

for this question nothing because i have no idea how to start it.

Okay, let's begin with a great tool used in physics called the free body diagram:

View attachment 5705

Okay, we wish to find the magnitude of the force $F$, which is equal in magnitude to $F_x$. We know the magnitude of $w$ (the weight vector) and we know $\theta$ where $\theta+\beta=\dfrac{\pi}{2}$...so can you use some trigonometry to relate $w$, $F_x$ and $\theta$?

 

[Evgeny.Makarov]

It seems to me that the figure would be clearer if $F_x$ and $F_y$ are renamed into $w_x$ and $w_y$ (and $w_x$ changes the direction). As I understand, $\vec{w}=\vec{w}_x+\vec{w}_y$. The component $\vec{w}_y$ of $\vec{w}$ is counterbalanced by the normal force $\vec{n}$ and $\vec{w}_x$ is counterbalanced by $\vec{F}$. But $\vec{F}$ does not have a $y$ component.