Question

An 8.31-m, 267-N uniform ladder rests against a smooth (frictionless) wall. The coefficient of static friction between the ladder and the ground is 0.582, and the ladder makes a 52.6? angle with the ground. A 928 N person is standing on the ladder a distance d from the bottom end of the ladder.
(a) Draw a free-body diagram of the ladder. (Turn in part (a) on the sheet provided in your packet.)
(b) How far up the ladder (distance d) can the person climb before the ladder begins to slip? Hint: Among other things, you may want to look at the sum of the torques about the point where the ladder touches the wall. Be careful with angles! ___________ m ( ± 0.02 m)

Answer 1

Answer:

2.88m

Explanation:

Length of the ladder = 8.31m

weight of ladder = 267N

coefficient of static friction (u) = 0.582

Angle(α) = 52.6°

Weight of person = 928N

Normal reaction on the ladder = weight of ladder + weight of person

= 267 + 928

= 1195N

Friction force(F) = uR

= 0.582*1195

= 695.49N

Torque due to friction = F(L/2)sin52.6°

= 695.49(8.31/2)sin52.6

= 2295.67N

Torque due to person = weight of person *sCos 52.6

=928*sCos52.6

=s* 563.64N

Torque due to ladder = weight of ladder * (L/2)Cos52.6

= 267*(8.31/2)Cos 52.6

= 673.81N

Torque(person) + Torque (ladder) = Torque (friction)

2295.67 = s*563.64 + 673.81

2295.67 - 673.81 = s*563.64

1621.86 = s*563.64

s = 1621.86/563.64

s = 2.8774

s = 2.88m

The person is at 2.88m from the end of the ladder

Answer 2

Torque is the rotational equivalent force. The person is at 2.88m from the end of the ladder.

What is torque?

Torque is the rotational equivalent force. It calculates about a point,

$Torque = Force \times \text{perpendicular distance between the point and and the force}$

The following data is been given

• Length of the ladder = 8.31m
• Weight of ladder = 267N
• Coefficient of static friction (u) = 0.582
• Angle(α) = 52.6°
• Weight of person = 928N

If we calculate the vertical forces,

$\sum F_y = 0$

Reaction force on the ladder, N

= weight of ladder + weight of the person

= 267 + 928

= 1195N

Also, the Friction force(F) at the base of the ladder can be written as,

$F = \mu N\\= 0.582\times 1195\\= 695.49N$

Further, calculating the torques about the point where the ladder touches the wall,

Torque due to the friction force

$\begin{aligned}T_F &= F \times \dfrac{L \times\ Sin(52.6)}{2}\\&= 695.49\times \dfrac{8.312 \times\ Sin(52.6)}{2}\\&= 2295.67N\end{aligned}$

Torque due to the weight of the person

$\rm T_w = \text{weight of person} \times s\ Cos(52.6)$

     $\rm =928\times s\ Cos52.6\\= 563.64s\ N$

Torque due to the weight of the ladder

$T_L = \text{Weight of ladder} \times \dfrac{L\times Cos52.6}{2}$

$T_L = 267 \times \dfrac{8.31\times Cos52.6}{2}$

$T_L = 673.81\rm\ N$

Now, if take the sum of torque at the point where the ladder touches the wall,

$T_F=T_W + T_L\\\\2295.67 = (s \timse 563.64) + 673.81\\\\2295.67 - 673.81 = s \timse 563.64\\\\1621.86 = s \timse 563.64\\\\s = \dfrac{1621.86}{563.64}\\\\s = 2.8774 \approx 2.88\rm\ m$

Hence, the person is at 2.88m from the end of the ladder.

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