Max Upward Force to Hold Bowling Ball on Wall

In summary: All the forces contribute to torque in principle, but you can eliminate some by choosing an axis through which the forces pass. E.g. if you choose the centre of the ball then you can ignore the normal forces because they have no torque about...?In summary, In order to hold a Bowling Ball in place so it doesn't roll off the window ledge and land on someone below, you use a wire to hook one of the holes and apply a vertical force. The coefficient of static friction between the ball and ledge is the same as the coefficient of static friction between the ball and wall, so you can exert a maximum upward force of 10.2 N without the ball rotating.
  • #1
mattbeatlefreak
34
2

Homework Statement


A 4.5-kg bowling ball is perched on a concrete ledge directly below your dorm room window, with the side of the ball opposite the holes touching the wall. Wanting to hold the ball in place so that it doesn't roll off and land on somebody, you manage to hook one of the holes with a wire and exert a purely tangential (and vertical) force on the ball. The coefficient of static friction between ball and ledge is the same as that between ball and wall, μs = 0.43. What is the maximum upward force you can exert so that the ball does not rotate and you lose your hold? Even though the ball has holes drilled in it, assume a uniform distribution of inertia.

Homework Equations


τ = Fr
F = ma
ƒ = μN

The Attempt at a Solution


I drew a diagram and wrote equations for net torque, net force in x direction, and net force in y direction.

τnet=T-ƒwallledge=0
Fxledge-Nwall=0
Fy=T-mg+Nledgewall=0

To find the maximum force, I set the frictional force from the wall = μ times the normal force from the ledge. Also, I set the frictional force from the ledge = μ times the normal force from the wall.

Substituting values:
T-.43Nledge-.43Nwall=0
.43Nwall-Nwall=0
T-44.1+Nledge+.43Nledge=0

Solving for Nwall gave a value of 0, which doesn't seem correct. Continuing through regardless, I finished with an answer of 10.2 N for the tension, which wasn't correct.
 
Physics news on Phys.org
  • #2
mattbeatlefreak said:
I set the frictional force from the wall = μ times the normal force from the ledge. Also, I set the frictional force from the ledge = μ times the normal force from the wall.
Did you really do that? A little unconventional perhaps?
 
  • #3
haruspex said:
Did you really do that? A little unconventional perhaps?
I was told to do that from a tutor I was seeking help from. Clearly there was some misthought or miscommunication.
 
  • #4
mattbeatlefreak said:
I was told to do that from a tutor I was seeking help from. Clearly there was some misthought or miscommunication.
What's the usual relationship between maximum frictional force and normal force?
 
  • #5
haruspex said:
What's the usual relationship between maximum frictional force and normal force?
ƒs = μs x FN
 
  • #6
mattbeatlefreak said:
ƒs = μs x FN
Yes, but what are those two forces exactly? Where do they act and in what directions in relation to each other?
 
  • Like
Likes mattbeatlefreak
  • #7
haruspex said:
Yes, but what are those two forces exactly? Where do they act and in what directions in relation to each other?
Since friction is along a surface, and normal force is perpendicular to the surface, the forces should be perpendicular to one another. So in this case, the max friction force (along wall) would equal the μs times the normal force from the wall? And same thing for the ledge?
 
  • #8
mattbeatlefreak said:
Since friction is along a surface, and normal force is perpendicular to the surface, the forces should be perpendicular to one another. So in this case, the max friction force (along wall) would equal the μs times the normal force from the wall? And same thing for the ledge?
Yes.
 
  • #9
haruspex said:
Yes.
I got 15 N for the tension, and this answer is correct. Thanks for all your help!
 
  • #10
Care to explain further? I'm stumped on this one too.
 
  • #11
Scigirl said:
Care to explain further? I'm stumped on this one too.
As with the initial post of a thread, if you have the same homework you need to post your attempt.
Have you drawn a diagram? What forces act on the ball?
 
  • #12
Fx=Normalforce (wall)-Friction (ledge)=0
Fy=Normal(ledge)+Tension-mg-friction(wall)=0

Solved for Y direction for Normal Force Fn= (mg-T)/0.57
Then plugged into T=mg-fs-Fn
 
  • #13
Scigirl said:
Fx=Normalforce (wall)-Friction (ledge)=0
Fy=Normal(ledge)+Tension-mg-friction(wall)=0
You have two equations but three unknowns. What do you need to do?
 
  • #14
haruspex said:
You have two equations but three unknowns. What do you need to do?
Do I need a third equation? f=μN?
 
  • #15
Scigirl said:
Do I need a third equation? f=μN?
You need a third equation, but I was assuming you were already using f=μN. (If not, you had five unknowns.)
In 2D statics problems, there are always three equations available. Exactly which three is a matter of choice, but the usual choice is horizontal linear force balance, vertical linear force balance, and ...?
 
  • #16
haruspex said:
You need a third equation, but I was assuming you were already using f=μN. (If not, you had five unknowns.)
In 2D statics problems, there are always three equations available. Exactly which three is a matter of choice, but the usual choice is horizontal linear force balance, vertical linear force balance, and ...?
Rotational force? Torque perhaps? Torque=rF
 
  • #17
Scigirl said:
Rotational force? Torque perhaps? Torque=rF
Yes.
 
  • #18
haruspex said:
Yes.

I'm not sure how to use torque in this situation, is the Net torque from the tension and the 2 frictions?
 
  • #19
Scigirl said:
I'm not sure how to use torque in this situation, is the Net torque from the tension and the 2 frictions?
All the forces contribute to torque in principle, but you can eliminate some by choosing an axis through which the forces pass. E.g. if you choose the centre of the ball then you can ignore the normal forces because they have no torque about that point.
 
  • #20
Στ=τT-2τf
Tr=2(μFn)r
T=2μ[(mg-T)/(1-μ)]
This is what I got using Net Torque and ΣFy
 
  • #21
Scigirl said:
Στ=τT-2τf
Tr=2(μFn)r
T=2μ[(mg-T)/(1-μ)]
This is what I got using Net Torque and ΣFy
Are the two normal forces the same? It is important to use different symbols for variables with potentially different values.
 

1. What is the maximum upward force required to hold a bowling ball on a wall?

The maximum upward force required to hold a bowling ball on a wall depends on the weight of the ball and the angle of the wall. This force can be calculated using the formula F = mgtanθ, where F is the upward force, m is the mass of the ball, g is the acceleration due to gravity, and θ is the angle of the wall.

2. Why is the angle of the wall important in determining the maximum upward force?

The angle of the wall affects the amount of force that needs to be applied to hold the ball in place. The steeper the angle of the wall, the greater the upward force needed to counteract the pull of gravity on the ball.

3. Can the maximum upward force be greater than the weight of the bowling ball?

Yes, the maximum upward force can be greater than the weight of the bowling ball. This is because the force needed to hold the ball on the wall is not only dependent on the weight of the ball, but also on the angle of the wall.

4. How does the surface of the wall affect the maximum upward force?

The surface of the wall can affect the maximum upward force needed to hold the bowling ball in place. A rougher surface may provide more friction, making it easier to hold the ball on the wall with a smaller upward force. A smoother surface may have less friction, requiring a greater upward force to keep the ball in place.

5. What other factors can affect the maximum upward force to hold a bowling ball on a wall?

In addition to the weight of the ball and the angle and surface of the wall, other factors that can affect the maximum upward force include the strength and stability of the wall, as well as any external forces acting on the ball, such as wind or vibrations.

Similar threads

Replies
3
Views
3K
  • Introductory Physics Homework Help
Replies
8
Views
3K
  • Introductory Physics Homework Help
Replies
5
Views
2K
  • Introductory Physics Homework Help
Replies
18
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
4K
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
10
Views
4K
  • Introductory Physics Homework Help
Replies
10
Views
3K
  • Introductory Physics Homework Help
Replies
10
Views
1K
Replies
12
Views
4K
Back
Top