Upward Force on Bench Support Leg: 990N

[Maigowai]

Hi, I am stuck on this question...


A student(850 N) sits on a 2 m long bench that weighs 625 N. The student is sitting 0.40 m from the
right end of the bench. What upward force is exerted by the right support leg of the bench?


please help!


My attempt:
(0.40m)(820N)+(1m)(625N) = 965N
however, the answer that is given is 990 rounded.

 

[berkeman]

Hi, I am stuck on this question...


A student(850 N) sits on a 2 m long bench that weighs 625 N. The student is sitting 0.40 m from the
right end of the bench. What upward force is exerted by the right support leg of the bench?


please help!


My attempt:
(0.40m)(820N)+(1m)(625N) = 965N
however, the answer that is given is 990 rounded.

Welcome to the PF.

Could you please post a sketch of the problem, showing how you arrived at your equation? Thanks.

 

[Maigowai]

Welcome to the PF.

Could you please post a sketch of the problem, showing how you arrived at your equation? Thanks.

So what i did was, divide the bench's own force by 2 because it's being supported by 2 legs, but i used the torque equation r x F to find the force exerted 0.4m from the right side...then added it together to find the overall force on the right leg.

however my work seems to be flawed somewhere :(

the image attached is what i used to form my equation.

 

[berkeman]

Hi, I am stuck on this question...


A student(850 N) sits on a 2 m long bench that weighs 625 N. The student is sitting 0.40 m from the
right end of the bench. What upward force is exerted by the right support leg of the bench?


please help!


My attempt:
(0.40m)(820N)+(1m)(625N) = 965N
however, the answer that is given is 990 rounded.

So what i did was, divide the bench's own force by 2 because it's being supported by 2 legs, but i used the torque equation r x F to find the force exerted 0.4m from the right side...then added it together to find the overall force on the right leg.

however my work seems to be flawed somewhere :(

the image attached is what i used to form my equation.

Notice that the units in your first equation are not correct... Newtons * meters doesn't equal Newtons again...

Your approach is basically right, but be careful how you ration the forces. The right support holds up half of the weight of the bench, because that weight is centered in the middle. The person's weight is carried more by the right support because the weight is closer to the right. If the weight were in the middle of the plank, the right support would see half of the person's weight. If the weight were directly over the right support, the right support would carry all of the weight. Can you see now how to use the torque equation to get the right answer?

 

[Maigowai]

Notice that the units in your first equation are not correct... Newtons * meters doesn't equal Newtons again...

Your approach is basically right, but be careful how you ration the forces. The right support holds up half of the weight of the bench, because that weight is centered in the middle. The person's weight is carried more by the right support because the weight is closer to the right. If the weight were in the middle of the plank, the right support would see half of the person's weight. If the weight were directly over the right support, the right support would carry all of the weight. Can you see now how to use the torque equation to get the right answer?

if (850N)(0.4m)+(625N)(1m) is how much weight the right side is holding up...how would you divide the answer by (m) and still get aprox 990N?

 

[berkeman]

if (850N)(0.4m)+(625N)(1m) is how much weight the right side is holding up...how would you divide the answer by (m) and still get aprox 990N?

That equation is incorrect overall. You will end up dividing by the length of the bench, or half the length of the bench, depending on how you set your equations up.

Also notice how if you multiply by 0.4m for the weight of the person, you are getting less than half of his weight, even though he is *closer* to the right side. What should you be doing differently about that?

 

[Maigowai]

That equation is incorrect overall. You will end up dividing by the length of the bench, or half the length of the bench, depending on how you set your equations up.

Also notice how if you multiply by 0.4m for the weight of the person, you are getting less than half of his weight, even though he is *closer* to the right side. What should you be doing differently about that?

i found a formula that works, it is (850N)(1.6m)+(625N)(1m) = (2m)(F)
but i don't understand 2 things:
1) why do we use 1.6m and not 0.4m?
2) why must we divide the answer by the length of the bench?

 

[I like Serena]

Welcome to PF, Maigowai!

i found a formula that works, it is (850N)(1.6m)+(625N)(1m) = (2m)(F)
but i don't understand 2 things:
1) why do we use 1.6m and not 0.4m?
2) why must we divide the answer by the length of the bench?

You are talking about a moment sum here, which has to be zero to prevent rotation of the bench.

In every statics problem you deal with 3 equilibrium equations.
1. Sum of vertical forces is zero, so there is no vertical movement
2. Sum of horizontal forces is zero, so there is no horizontal movement
3. Sum of moment with respect to an arbitrary point is zero, so there is no rotation.

Are you familiar with moments?

A moment is a force times the distance of that force to the reference point.
These are the products that you have been showing.

In your case you need the 3rd equation, the moment sum with respect to the left leg of the bench.

 

[tiny-tim]

Welcome to PF!

Hi Maigowai! Welcome to PF!

1) why do we use 1.6m and not 0.4m?
2) why must we divide the answer by the length of the bench?

You are taking https://www.physicsforums.com/library.php?do=view_item&itemid=64" about the left end of the bench.

(you could do it about any point, but then you'd need to include the moment of the left https://www.physicsforums.com/library.php?do=view_item&itemid=73" … if you take moments about the point of application of that force, then of course it has zero moment, ie you can forget about it )

The sum of the moments has to be zero.

So the clockwise moments must equal the anti-clockwise moments … this equation puts the clockwise moments on the LHS, and the anti-clockwise moments on the RHS.

So (on the LHS), the distance from the left to the person is 1.6 m.

And (on the RHS), the distance to the reaction force is 2.0 m.