Forces necessary to hold a board steady (double check my work?)

In summary, you set the variables fx, fy, and B equal to zero in an attempt to hold the board still. However, because the net force is zero, each individual component of the force is also zero. Using the third equation, you solved for B and got -477.4. Plugging B into the first two equations, you then calculated Fy=854.4 and Fx=-238.7.
  • #1
Lotus93
6
0
This is the problem:
http://oi47.tinypic.com/6qij4o.jpg

I set fx, fy, and B equal to zero.
'x' forces: Fx - Bsin(30) = 0
'y' forces: Fy - mg + B(cos(30)) = 0
lower end: -0.5*mg(cos(30)) + 0.4*B = 0

Using the third equation I solved for B, and got -477.4.
I then plugged B into the first two equations and calculated Fy=854.4 and Fx=-238.7.

I don't feel confident in these numbers, so I wanted to double check with someone who has a better understanding. Did I go wrong somewhere, or does this look okay?
Thanks in advance, I really appreciate your help.
 
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  • #2
Lotus93 said:
This is the problem:
http://oi47.tinypic.com/6qij4o.jpg

I set fx, fy, and B equal to zero.
Why set all those to zero?
'x' forces: Fx - Bsin(30) = 0
'y' forces: Fy - mg + B(cos(30)) = 0
lower end: -0.5*mg(cos(30)) + 0.4*B = 0

Using the third equation I solved for B, and got -477.4.
I then plugged B into the first two equations and calculated Fy=854.4 and Fx=-238.7.

I don't feel confident in these numbers, so I wanted to double check with someone who has a better understanding. Did I go wrong somewhere, or does this look okay?
Thanks in advance, I really appreciate your help.
B should be positive !
 
  • #3
I set them equal to zero because I thought the net force should be equal to 0 in order to hold the board still... is that wrong?
 
  • #4
This is what I have now...

(-0.5)(m)(g)(cos(30))+(B)(0.4)=0
B=-(-.5)(45)(9.8)(cos(30))/0.4
B= 477.4

Fx-B(sin(30))=0
Fx=(477.4)(sin(30))
Fx= 238.7

Fy-mg+B(Cos(30))=0
Fy=-477.4(Cos(30))+(45)(9.8)
Fy=27.6Am I anywhere close to being right?
 
Last edited:
  • #5
Lotus93 said:
I set them equal to zero because I thought the net force should be equal to 0 in order to hold the board still... is that wrong?
The net force is zero, but that doesn't mean that each individual component making up the net force is zero.
 
  • #6
Lotus93 said:
This is what I have now...

(-0.5)(m)(g)(cos(30))+(B)(0.4)=0
B=-(-.5)(45)(9.8)(cos(30))/0.4
B= 477.4

Fx-B(sin(30))=0
Fx=(477.4)(sin(30))
Fx= 238.7

Fy-mg+B(Cos(30))=0
Fy=-477.4(Cos(30))+(45)(9.8)
Fy=27.6

Am I anywhere close to being right?
The method is right.

The numbers look reasonable.

Of course, all of those answers are in units of Newtons.
 

1. What are the forces involved in holding a board steady?

The main forces involved in holding a board steady are friction, normal force, and gravitational force. Friction acts in the opposite direction of the applied force, while normal force is perpendicular to the surface of the board and counteracts the weight of the object.

2. How does the weight of the object affect the necessary forces to hold the board steady?

The weight of the object plays a significant role in determining the necessary forces to hold the board steady. The greater the weight of the object, the larger the normal force and frictional force must be to keep the board from slipping or toppling.

3. What factors can influence the frictional force required to hold a board steady?

The coefficient of friction, the weight of the object, and the type of surface the board is resting on can all influence the frictional force needed to hold the board steady. A higher coefficient of friction and a heavier object will require a greater frictional force, while a rougher surface can increase the frictional force as well.

4. How can the angle of the board affect the necessary forces to hold it steady?

The angle of the board can significantly impact the necessary forces to hold it steady. As the angle of the board increases, the normal force and frictional force must also increase to prevent the object from sliding or falling off the board. This is because the weight of the object is acting more parallel to the surface, making it easier for it to overcome the frictional force.

5. Are there any other forces that can affect the stability of a board?

Aside from the main forces of friction, normal force, and gravitational force, other external forces such as wind or vibrations can also impact the stability of a board. Additionally, if the board is not properly supported or if there are any defects in the board itself, these factors can also affect its ability to remain steady.

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