Ask question

Ask question

All categories

vladimir2022 [97]

3 years ago

6

A uniform log of length L is inclined 30° from the horizontal when supported by a frictionless rock located 0.6L from its left e

nd. The mass of the log is 200 kg. An engineer with a mass of 53.5 kg walks along the log from the left to the right until the log is balanced horizontally. How far from the left end of the log is the engineer when the log is horizontal?

1 answer:

mafiozo [28]3 years ago

7 0

Answer:

x = 0.974 L

Explanation:

given,

length of inclination of log = 30°

mass of log = 200 Kg

rock is located at = 0.6 L

L is the length of the log

mass of engineer = 53.5 Kg

let x be the distance from left at which log is horizontal.

For log to be horizontal system should be in equilibrium

∑ M = 0

mass of the log will be concentrated at the center

distance of rock from CM of log = 0.1 L

now,

∑ M = 0

$m_{log} g \times 0.1 L = m_{engineer} g \times (x - 0.6 L)$

$200 \times 0.1 L = 53.5 \times (x - 0.6 L)$

$0.374 L =x - 0.6 L$

x = 0.974 L

hence, distance of the engineer from the left side is equal to x = 0.974 L

You might be interested in

Heres a random question just for fun and out. of boredom ok so who started the nasa program? and why? what made him want to do t

Pachacha [2.7K]

Eisenhower started the NASA project to develop technology for military application.

5 0

3 years ago

4.1 A steel spur pinion has a pitch of 5 teeth/in, 20 full-depth teeth, and a 20° pressure angle. The pinion runs at a speed of

Vesnalui [34]

Answer:

15.07 ksi

Explanation:

Given that:

Pitch (P) = 5 teeth/in

Pressure angle ( $\Phi$ ) = 20°

Pinion speed ( $n_p$ ) = 2000 rev/min

Power (H) = 30 hp

Teeth on gear ( $N_G$ ) = 50

Teeth on pinion ( $N_p$ ) = 20

Face width (F) = 1 in

Let us first determine the diameter (d) of the pinion.

Diameter (d) = $\frac{N}{P}$

= $\frac{20}{5}$

= 4 in

From the values of Lewis Form Factor Y for ( $n_p$ ) = 20 ; at 20°

Y = 0.321

To find the velocity (V); we use the formula:

$V = \frac{\pi d n_p}{12}$

$V = \frac{\pi *4*2000}{12}$

V = 2094.40 ft/min

For cut or milled profile; the velocity factor $(K_v)$ can be determined as follows:

$K_v = \frac{2000+V}{2000}$

$K_v = \frac{2000+2094.40}{2000}$

= 2.0472

However, there is need to get the value of the tangential load $(W^t)$ , in order to achieve that, we have the following expression

$W^t=\frac{T}{\frac{d}{2} }$

$W^t = \frac{63025*H}{\frac{n_pd}{2}}$

$W^t = \frac{63025*30}{2000*\frac{4}{2}}$

$W^t = 472.69 lbf$

Finally, the bending stress is calculated via the formula:

$\sigma = \frac{K_vW^tp}{FY}$

$\sigma = \frac{2.0472*472.69*5}{1*0.321}$

$\sigma = 15073.07 psi$

$\sigma =$ 15.07 ksi

∴ The estimate of the bending stress = 15.07 ksi

4 0

3 years ago

Read 2 more answers

Two objects are dropped from rest from the same height. Object A falls through a distance Da and during a time t, and object B f

stiv31 [10]

Answer:

Da=(1/4)Db

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

When s = Da, t = t

$s=ut+\frac{1}{2}at^2\\\Rightarrow Da=0\times t+\frac{1}{2}\times a\times t^2\\\Rightarrow Da=\frac{1}{2}at^2$

When s = Db, t = 2t

$s=ut+\frac{1}{2}at^2\\\Rightarrow Da=0\times t+\frac{1}{2}\times a\times (2t)^2\\\Rightarrow Db=\frac{1}{2}a4t^2$

Dividing the two equations

$\frac{Da}{Db}=\frac{\frac{1}{2}at^2}{\frac{1}{2}a4t^2}=\frac{1}{4}\\\Rightarrow \frac{Da}{Db}=\frac{1}{4}\\\Rightarrow Da=\frac{1}{4}Db$

Hence, Da=(1/4)Db

3 0

3 years ago

Real life example for each gas law

Dvinal [7]

Applications of Gas Law in Real Life. A torch used to heat up the and rise the air temperature inside the balloon. This cause the air volume inside the balloon to increased and becoming less dense than the surrounding air. ... The air in the ears will change its volume then causes yours ears to pop due to the strain.

3 0

3 years ago

In a fluorescent tube of diameter 4.8 cm , 2.7 × 1018 electrons (with a charge of −e) and 2.4 × 1018 positive ions (with a charg

Lesechka [4]

Answer: 0.817A

Explanation:

Assuming , that one coulomb per second of negative charge alone flow through a conductor and no positive charges flow. I.e Q=It

It means a current of one A flow in the opposite direction.

This is similar to one coulomb per second of positive charge flowing through and there is no negative charge,

In addition, the one coulomb per second of positive charge flows. This is flowing in the current direction of the previous one. Then, the total current is 2 A. Since 2 coulomb of positive charges flow through one due to real positive charge and another due to the negative charge flowing in opposite direction.The charges cannot cancel each other, because even before the current flow the conductor was neutral.

According to this, the current in the given problem is

[2.7 + 2.4] x 10 ^ 18 * 1.602 x 10^ [-19] C/s

= 0.817 A

7 0

3 years ago