Finding the Force on a bolt from a wrench

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    Bolt Force Wrench
AI Thread Summary
The discussion centers on calculating the force applied to a bolt when tightening it with a wrench. Participants analyze the torque equations and the relationship between the wrench length and bolt radius. There is confusion regarding the dimensions of the answers provided, as none correspond to torque, leading to a debate about the correct interpretation of the problem. The consensus suggests that the force exerted on the bolt's thread is related to the torque produced by the wrench, emphasizing the importance of understanding the mechanical principles involved. Ultimately, the conversation highlights the need for clarity in problem statements and the significance of torque in such engineering scenarios.
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Homework Statement


In an Effort to tighten a bolt, a force F is applied to the end of a wrench that has a length L perpendicular to the wrench. If the bolt itself has a radius of b, how much force is applied to the bolt

Homework Equations


∑τ=RFsinθ ΣF=ma ∑T=Iα

The Attempt at a Solution


I tried summing the torques
∑τ=LF+bF=(I_bolt+I_wrench)α

but the answer choices are
a)F
b)Fb/L
c)FL/b
d)FL/(b+L)
e)bL/F

I'm not even sure how you would solve this to get anything like that or why there is an F in the answer when we're solving for F. My friend says this is a conceptual engineering problems and that c makes sense since the Force is multiplied by the length of the bolt (they are really looking for torque) but divided by the length of the wrench.
 
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pinkfishegg said:
how much force is applied to the bolt
That's not the title of your thread. You sure they aren't out to trick you ?

And: make a drawing
 
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here's a pic drawn in paint
 

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  • wrench physics.png
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L is a bit vague, don't you think ?

And do you think people mean that when they mention the radius of a bolt ? In the civilized world the M10 bolt has a radius of 5 mm, but the head is a bit bigger !

pinkfishegg said:
2. Homework Equations
∑τ=RFsinθ
ΣF=ma
∑T=Iα

3. The Attempt at a Solution
I tried summing the torques
∑τ=LF+bF=(I_bolt+I_wrench)α

Note that there is only one F in your drawing !

You be the judge: answering the question literally is a). But the force the thread of the bolt exerts on the thread on the inside of the hole (or nut) is c) as your friend sort of indicated. This multiplication factor is the reason wrenches exist :smile:

(Note that the length of a bolt is in general not the same as the radius...)

You know the torque from L x F. (angle is 90 degrees). At the thread of the bolt the torque is the same !

None of the answers has the dimension of torque ( Newton x meter or better: Force x length). All of them have the dimension of a force.
 
So are we looking for a ratio of the torques to find the Force?
 
BvU said:
L is a bit vague, don't you think ?

And do you think people mean that when they mention the radius of a bolt ? In the civilized world the M10 bolt has a radius of 5 mm, but the head is a bit bigger !
Note that there is only one F in your drawing !

You be the judge: answering the question literally is a). But the force the thread of the bolt exerts on the thread on the inside of the hole (or nut) is c) as your friend sort of indicated. This multiplication factor is the reason wrenches exist :smile:

(Note that the length of a bolt is in general not the same as the radius...)

You know the torque from L x F. (angle is 90 degrees). At the thread of the bolt the torque is the same !

None of the answers has the dimension of torque ( Newton x meter or better: Force x length). All of them have the dimension of a force.

Oh a radius of a bolt is the round thing before the threat, is the force applied there?
 

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  • radiusbolt.jpg
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No. That's a specific type of bolt and that radius has little to do with the exercise.

I meant the radius of the shaft: so D/2 in the drawing below

bolts.gif


pinkfishegg said:
So are we looking for a ratio of the torques to find the Force?
I tried to explain that there is only one torque and that the ratio of the forces at L and at b is the ratio of b and L : ##\tau = L \times F_L ## and ##\tau = b \times F_b##
 
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BvU said:
No. That's a specific type of bolt and that radius has little to do with the exercise.

I meant the radius of the shaft: so D/2 in the drawing below

bolts.gif


I tried to explain that there is only one torque and that the ratio of the forces at L and at b is the ratio of L and b : ##\tau = L \times F_L ## and ##\tau = b \times F_b##

Ohhh that makes so much sense thanks
 
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