How Much Force is Needed to Move a Block Against Friction and Gravity?

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The discussion focuses on calculating the minimum force required to move a block against friction and gravity when it is in contact with a vertical wall. The weight of the block is 88.9N, and the coefficient of static friction is 0.560. To prevent the block from sliding down, the minimum force required is calculated to be approximately 88.2N, while the force needed to start moving the block upwards is around 228N. Participants emphasize the importance of correctly identifying the normal force and applying the appropriate equations for static friction. The conversation reveals confusion over free body diagrams and the application of forces at angles, highlighting the need for clarity in problem-solving.
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1. There is a block parallel and touching a vertical wall, a force F is applied at a 40 deg. angle from -x,-y. the weight of the block is 88.9N. The coefficient of static friction between the block and the wall is 0.560.
a) what is the minimum force F required to prevent the block from sliding down the wall?
b) what is the minimum force F required to start the block moving up the wall?



Homework Equations





3.a) Fs(max)=(0.560)(88.9N)=49.8N
Fy=88.9N-49.8N=39.1N
Fx=(39.1N)/(cos(40))=51.0N
\sqrt{39.1^2+51.0^2}=F
F=64.3N

b) 88.9N +49.8 N=138.7N=Fy
Fx=(138.7N)/(cos(40))=181.06N
\sqrt{138.7^2+181.06^2}=F
F=228N
 
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You should be more specific...

I saw an error on your equations. The friction equation is defined as F_{sf}=\mu_{s}*N.

N is the normal force of the surface. In your case is the Normal force from the wall and you used the weight of the box. You should use as N the x component of the force F you should apply.

Try again knowing this fact.

Note: Superscript should be underscript.
 
a)88.9N*tan(40)=74.6N=Fn
Fy=88.9N
Fs=(0.560)*(74.6N)=41.8N
88.9N-41.8N=47.1N
\sqrt{47.1^2+74.6^2}=88.2N
F=88.2N
 
mslena79 said:
a)88.9N*tan(40)=74.6N=Fn
Fy=88.9N
Fs=(0.560)*(74.6N)=41.8N
88.9N-41.8N=47.1N
\sqrt{47.1^2+74.6^2}=88.2N
F=88.2N

Are you guessing?

Here, the least force applied to prevent motion is when friction vector is pointing upward.
Write both of your equations, \Sigma\vec{F}=0 x and y.

Solve equations simultaneously. You unknowns are N and F and you have two equations.
 
I don't understand.
 
mslena79 said:
I don't understand.

Due to my english? It can't be that bad.
 
No, I think I am having trouble because I am not sure where to place the applied force vector on my free body diagram. I have fs pointing +y, mg -y, Fn=-x, F 40 deg. in the (-x,-y)quadrant.
 
mslena79 said:
No, I think I am having trouble because I am not sure where to place the applied force vector on my free body diagram. I have fs pointing +y, mg -y, Fn=-x, F 40 deg. in the (-x,-y)quadrant.

Ahh. That's they way it's suppose to be. Now you make use the following equations

\sum Fx=0

-Fn+F*cos(40)=0

Note: I assumed the angle given is the angle between the force and the x axis, otherwise you should use sin

\sum Fy=0

Fs-m*g+F*sin(40)=0 Fs=\mu s*Fn so first equation for Fn you should have:

Fn=F*cos(40) and you substitute this equation on the \sum Fy=0 equation:

F*cos(40)-m*g+F*sin(40)=0 and then F=\frac{m*g}{cos(40)+sin(40)}

;-)
 
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I recently came across this question while looking for solutions to mine (which was essentially the same thing). After solving it, I thought that I'd come back and add on to the solution, in case there were still some questions or somebody else is in need of assistance.

Fcos(x)-w+usFn=0 and Fsin(x)-Fn=0

After simplifying the two equations that, together, find the solution to the problem, one ends up with:

\frac{W}{cos(\Theta\degree)+\mu_{s}sin(\Theta\degree)}=F_{n}

where Theta is the angle measurement, Mu is the coefficient, and W is the weight (or force).
 
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Likewise, for the second part, simply subtract the second term in the denominator rather than adding, because the frictional force is in the opposite direction.
 

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