As shown in the figure, a beam is supported by two pillars separated by a distan
ID: 1465670 • Letter: A
Question
As shown in the figure, a beam is supported by two pillars separated by a distance ?. The beam has a mass M, a length L, and a girl of mass m is walking from the left end toward the right.
(a) Find a symbolic expression for the normal force exerted on the beam by the pillar on the right, when the beam is on the verge of tipping. (Use the following as necessary: M, m, and g.)
(b) Find a symbolic expression for the girl's position, when the beam is on the verge of tipping. (Use the following as necessary: M, m, L and ?.)
(c) Find a symbolic expression for the minimum value of ? that will allow the girl to reach the end of the beam without it tipping. (Use the following as necessary: M,m, and L.)
Explanation / Answer
a) When the beam is on the verge of tipping, there is yet no motion, so the sum of the torques is still equal to zero. All we need to do to solve this problem is write the sum of the forces on the beam and equate this to zero. Keep in mind that when the beam is about to tip, the normal force on the first pivot is equal to zero. But, because it makes the solution more general and simultaneously solves parts a and b, we will instead write the sum of the torques about the left pivot and equate this to zero.
Taking up to be positive and measuring torques about the left pivot:
N*l - m*g*x - M*g*(L/2) = 0
Where N is the normal force exerted by the second pivot and x is the distance of the woman from the first pivot. M*g is multiplied by L/2 because the force of gravity acts on the center of mass, which is clearly at L/2. The obvious assumption is that L>l. A less obvious assumption is that l>L/2, otherwise the beam would tip before the woman stepped onto it.
We now have two unknowns, x and N. We therefore need another expression relating the two to solve for both. To do this, we write the sum of the torques about the second pivot, which is also equal to zero for the same reason the sum of the torques about the first is.
M*g*(l - L/2) - m*g*(x-l) = 0
Solving for N and x, we obtain the following results (I won't do the math here, but it wouldn't hurt to check my work):
N = (m + M)*g
x = (M/m)*(l - L/2) + l = (M/m)*[(M*l+m*l)/M - L/2]
b) x = (M/m)*(l - L/2) + l = (M/m)*[(M*l+m*l)/M - L/2]
c) Set x from part b equal to L, the end of the beam, and solve for l:
L = (M/m)*[(M*l+m*l)/M - L/2]
l = L*(m + M/2)/(m + M)
The technique used to solve part a can be used to solve any balance-beam type problem. The problem given is a special case when the first normal force is equal to zero. Generality is a goal in all of physics, which is why I suggest using the above method as practice. Alternatively, the answer to part a can be obtained by equating the forces only:
N - m*g - M*g = 0
N = g*(m + M)
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