Solve Statics Problem: Find Force Compressing A & B

[walnuts]

Homework Statement

Figure 12-61 is an overhead view of a rigid rod that turns about a vertical axle until the identical rubber stoppers A and B are forced against rigid walls at distances rA = 7.0 cm and rB = 3.2 cm from the axle. Initially the stoppers touch the walls without being compressed. Then force of magnitude 110 N is applied perpendicular to the rod at a distance R = 5.6 cm from the axle. Find the magnitude of the force compressing (a) stopper A and (b) stopper B.

The distance R is between the axle and stopper A.

Homework Equations

torque=(radius)*(force)
torque(net)=0

The Attempt at a Solution

For statics I usually pick a pivet point, but since there is an axle will I have to use that as reference?

using the stopper at rB as the pivet point.
Fb= force on stopper b
Fa=force on stopper a
F=applied force

Fb*(rB) + F*(rB+R)-Fa*(rA+rB)=0

F*(rB+R)=Fa*(rA+rB)

(F*(rB+R))/(rA+rB)=Fa

Fa-F=Fb

and it's wrong. [/b]

 

[anathema]

Hello,

Thank you for your question. First, let's clarify some of the given information. The figure shows a rigid rod, which means it does not deform under applied forces. The stoppers A and B are identical, and are initially touching the walls without compression. This means that their initial distance from the axle (rA and rB) is equal to the distance between the axle and the walls. Finally, the force applied is perpendicular to the rod, meaning it is acting at a 90 degree angle to the rod.

Now, let's move on to solving the problem. Since the rod is not deforming, we can apply the principle of torque equilibrium, which states that the net torque acting on a system is equal to zero. In this case, we can choose any point as our pivot point, as long as we are consistent with our choice throughout the problem.

Let's choose the pivot point to be the axle itself. This means that the distance from the pivot point to stopper A (rA) and stopper B (rB) are equal to 7.0 cm and 3.2 cm, respectively. We also know that the applied force (F) is acting at a distance of 5.6 cm from the pivot point. Now, we can apply the torque equation:

torque(net) = 0

(rA * Fa) + (rB * Fb) - (R * F) = 0

We can simplify this equation by substituting the value of Fa and Fb from the given information:

(rA * F) + (rB * (F - Fa)) - (R * F) = 0

Now, we can solve for Fa and Fb:

(rA * F) + (rB * F) - (rB * Fa) - (R * F) = 0

(rA * F) + (rB * F) - (rB * ((F * (rB + R)) / (rA + rB))) - (R * F) = 0

(rA * F) + (rB * F) - ((rB * F * rB) + (rB * F * R)) / (rA + rB) - (R * F) = 0

(rA * F) + (rB * F) - ((rB * F * rB) + (

 

[kerimek]

I would approach this problem by first defining the variables and parameters involved. In this case, the given information includes the distances rA, rB, and R, and the applied force of magnitude 110 N. The unknowns are the forces Fa and Fb, which represent the force compressing stoppers A and B, respectively.

Next, I would use the equations for torque to set up an equation that relates the forces and distances involved. Since the stoppers are forced against rigid walls, the net torque must be zero. This leads to the equation:

Fb(rB) + F(rB+R) - Fa(rA+rB) = 0

Solving for Fa, we get:

Fa = (F(rB+R))/(rA+rB)

Substituting this expression for Fa into the equation for the net torque, we can solve for Fb:

Fb = Fa - F = (F(rB+R))/(rA+rB) - F

Plugging in the given values for rA, rB, R, and F, we can calculate the magnitude of the force compressing stoppers A and B.

Fa = (110 N)(3.2 cm + 5.6 cm)/(7.0 cm + 3.2 cm) = 110 N (9.2 cm)/(10.2 cm) = 99.8 N

Fb = 99.8 N - 110 N = -10.2 N

Note that the negative sign for Fb indicates that the force is acting in the opposite direction of Fa. This means that stopper B is being pulled away from the rigid wall, while stopper A is being pushed towards it.

In conclusion, the force compressing stoppers A and B are 99.8 N and -10.2 N, respectively. This solution assumes that the rod and stoppers are in equilibrium, with no other external forces acting on them.