physics help please

Question

A beam resting on two pivots has a length of
L = 6.00 m
and mass
M = 87.0 kg.
The pivot under the left end exerts a normal force
n1
on the beam, and the second pivot placed a distance
l = 4.00 m
from the left end exerts a normal force
n2.
A woman of mass
m = 54.0 kg
steps onto the left end of the beam and begins walking to the right as in the figure below. The goal is to find the woman's position when the beam begins to tip.
(b) Where is the woman when the normal force
n1
is the greatest?
x = m

(c) What is
n1
when the beam is about to tip?
N

(d) Use the force equation of equilibrium to find the value of
n2
when the beam is about to tip.
N

(e) Using the result of part (c) and the torque equilibrium equation, with torques computed around the second pivot point, find the woman's position when the beam is about to tip.
x = m

(f) Check the answer to part (e) by computing torques around the first pivot point.
x = m

part 2
A beam of length L and mass M rests on two pivots. The first pivot is at the left end, taken as the origin, and the second pivot is at a distance l from the left end. A woman of mass m starts at the left end and walks toward the right end as in the figure below.

(a) When the beam is on the verge of tipping, find a symbolic expression for the normal force exerted by the second pivot in terms of M, m, and g. (Enter the magnitude.)
n2 =


(b) When the beam is on the verge of tipping, find a symbolic expression for the woman's position in terms of M, m, L, and l.
x =


(c) Find the minimum value of l that will allow the woman to reach the end of the beam without it tipping. (Use the following as necessary: m, M, and L.)
lmin =

i know its a loaded question but I'm super stuck. please help me

Answer 1

The beam begins to tip when the normal force at the left pivot (𝑛₁) reaches zero. At this point, the beam pivots about the second pivot. Determine the woman's position (𝑥) at the onset of tipping by considering torque equilibrium around the second pivot. Simultaneously, calculate the normal force (𝑛₂) using vertical force balance at the tipping point.

(b) The woman is farthest left when 𝑛 1 is greatest.

𝑥=0 m

(c) When the beam is about to tip, the left reaction force, n₁= 0 N.

(d) Determine n2 when the beam is about to tip using force equilibrium. At the tipping point, n1 = 0. Therefore, from the vertical force balance, n2 = (M + m)g = (87.0 + 54.0)(9.80) = 1381.8 N.

n_2​=1381.8 N

Part (e) — torque about second pivot (find 𝑥 when tipping) Take moments about the second pivot (distance 𝑙 from the left end). Sum of torques = 0 (clockwise positive):

Woman’s torque: 𝑚 𝑔 ( 𝑥 − 𝑙 ) . Beam’s self-weight: 𝑀 𝑔 acting at beam center 𝐿 / 2 gives torque 𝑀 𝑔 ( 𝐿 / 2 − 𝑙 ) .

Set their algebraic sum to zero:

𝑚 𝑔 ( 𝑥 − 𝑙 ) + 𝑀 𝑔 ( 𝐿 /2 − 𝑙 )

1. 
   

. Cancel 𝑔 and solve for 𝑥 .

4.00 − 87.0/ 54.0 ( 3.00 − 4.00 )

=4.00 − 87/ 54 ( − 1.00 )

=4.00 + 87/ 54

4.00 + 1.6111 …

=5.6111 m .

(so tipping happens before she reaches the far end at 6.00 m).

(f) Verification by torque calculation: Summing torques about the left pivot (at x=0), the upward force $n_2$ at $x=l$ balances the downward weights, such that $n_2 l - Mg(L/2) - mgx = 0$. Solving for $n_2$ yields $n_2 = \frac{Mg(L/2) + mgx}{l}$. Substituting the given values, we find $n_2 = \frac{9.80(87 \cdot 3.00 + 54 \cdot 5.6111)}{4.00} = 9.80 \cdot 141.0 = 1381.8 \text{ N}$, which agrees with the result in part (d). ✔