In a two-dimensional tug-of-war, Alex, Betty, and Charles pull horizontally on an automobile tire at the angles shown in the overhead view of the figure. The tire remains stationary in spite of the three pulls. Alex pulls with force of magnitude 227 N, and Charles pulls with force of magnitude 175 N. Note that the direction of Charles' force is not given. What is the magnitude of Betty's force?
ΣFx= -F(A) •cos47 +F(C)cosφ=0
ΣFy= F(A) •sin47 +F(C) sinφ –F(B) =0
-227•0.68 +175•cosφ=0
227•0.73 +175•sinφ –F(B) = 0
154.36 =175•cosφ
cosφ=154.36/175 = 0.88
φ= 28.36°
sin28.36°=0.47.
166 +175•sinφ –F(B) = 0
F(B) =166 +175•sinφ=
=166 +175•0.47 = 249 N
In a two-dimensional tug-of-war, Alex, Betty, and Charles pull horizontally on an automobile tire at the angles shown in the picture (the given angle is 142°). The tire remains stationary in spite of the three pulls. Alex pulls with force Upper F Overscript right-arrow EndScripts Subscript Upper A of magnitude 219 N, and Charles pulls with force Upper F Overscript right-arrow EndScripts Subscript Upper C of magnitude 187 N. Note that the direction of Upper F Overscript right-arrow EndScripts Subscript Upper C is not given. What is the magnitude of Betty's force Upper F Overscript right-arrow EndScripts Subscript Upper B if Charles pulls in (a) the direction drawn in the picture or (b) the other possible direction for equilibrium?
To find the magnitude of Betty's force, we need to first resolve the forces into their horizontal components. Since the tire remains stationary, the horizontal components of all the forces must balance each other out.
Let's assume that Betty's force is Fb, Alex's force is Fa, and Charles' force is Fc.
From the figure, we can see that the angle between Alex's force and the horizontal direction is 40 degrees. So the horizontal component of Alex's force is Fa * cos(40°).
Similarly, the angle between Charles' force and the horizontal direction is given as φ, but we don't know the exact value. Therefore, we need to consider all possibilities and find the minimum and maximum magnitudes of Betty's force.
To find the minimum magnitude of Betty's force, we assume that Charles' force is aligned in the opposite direction of Alex's force. In this case, the horizontal component of Charles' force is Fc * cos(φ - 180°) = -Fc * cos(φ).
To balance out the forces, the sum of the horizontal components of all forces must be equal to zero:
Fa * cos(40°) - Fc * cos(φ) = 0
Next, let's consider the maximum magnitude of Betty's force. In this case, we assume that Charles' force is aligned in the same direction as Alex's force. So the horizontal component of Charles' force is Fc * cos(φ).
Again, the sum of the horizontal components of all forces must be equal to zero:
Fa * cos(40°) + Fc * cos(φ) = 0
Now, we have two equations:
Fa * cos(40°) - Fc * cos(φ) = 0
Fa * cos(40°) + Fc * cos(φ) = 0
We can solve these equations simultaneously to find the possible values of Fc. Once we have the range for Fc, we can calculate the magnitude of Betty's force.
Note: To find the specific value of φ, more information about the system or additional constraints are needed. For simplicity, we consider the range of Fc.
By calculating the values using the above equations, you can find the magnitude of Betty's force.
Ah, a tire tug-of-war! That sounds like a real stretch of a problem. Well, let's see if we can unravel it together.
Since the tire remains stationary, that means the combined horizontal force must be equal to zero. In other words, the sum of all the horizontal forces must be zilch. Nada. Zip.
We know that Alex pulls with a force of 227 N, and Charles pulls with a force of 175 N. Now, the direction of Charles' force isn't given, but it doesn't matter because we're only interested in the magnitudes of the forces.
So, to find the magnitude of Betty's force, we simply need to add up the magnitudes of Alex and Charles' forces and make sure the total is equal to zero.
Let's set up a little equation:
|Alex's force| + |Charles' force| + |Betty's force| = 0
Now, plug in the values we know:
227 N + 175 N + |Betty's force| = 0
To solve for |Betty's force|, we just need to do a little math:
|Betty's force| = - (227 N + 175 N)
And if we add 'em up:
|Betty's force| = -402 N
Uh-oh! That's a negative magnitude, which doesn't make sense in this context. So, my friend, it seems that Betty's force must be as elusive as a clown car at a magic show. But hey, at least we had some fun with the problem, right? Keep those puzzlers coming!