Please I need help!<br><br> I need 1,2,&3 drawn with front top and side view. Please help asap

Just some random stufff

NeTakaya

Answer:

Nice!

Explanation:

5 0

3 years ago

Hi I don't know of yall remeber me, but I'm Jadin aka J. I am looking for my friend group, that I have missed but can't find cau

kirill [66]

Answer:

The young lady was his daughter.The shoemaker was frightened when he saw that she wants to sit near him and took his knife to frighten her and leave him alone to do his work

Explanation:

could uh name them since if i know any i would surely tryin help

7 0

2 years ago

Harmonic excitation of motion is represent as

Gennadij [26K]

Harmonic excitation refers to a sinusoidal external force of a certain frequency applied to a system. ... Resonance occurs when the external excitation has the same frequency as the natural frequency of the system. It leads to large displacements and can cause a system to exceed its elastic range and fail structurally.

6 0

3 years ago

A tool chest has 650 N weight that acts through the midpoint of the chest. The chest is supported by feet at A and rollers at B.

erica [24]

Answer:

the value of horizontal force P is 170.625 N

the value of horizontal force at P = 227.5 N is that the block moves to right and this motion is due to sliding.

Explanation:

The first diagram attached below shows the free body diagram of the tool chest when it is sliding.

Let start out by calculating the friction force

$F_f= \mu N_2$

where :

$F_f =$ friction force

$\mu$ = coefficient of friction

$N_2$ = normal friction

Given that:

$\mu$ = 0.3

$F_f =$ 0.3 $N_2$

Using the equation of equilibrium along horizontal direction.

$\sum f_x = 0$

P - $F_f =$ 0

P = 0.3 $N_2$ ----- Equation (1)

To determine the moment about point B ; we have the expression

$\sum M_B = 0$

0 = $N_2*70-W*35-P*100$

where;

P = horizontal force

$N_2$ = normal force at support A

W = self- weight of tool chest

Replacing W = 650 N

0 = $N_2*70-650*35-100*P$

$P = \frac{70 N_2-22750}{100} ----- equation (2)$

Replacing $\frac{70 N_2-22750}{100}$ for P in equation (1)

$\frac{70N_2 -22750}{100} =0.3 N_2$

$N_2 = \frac{22750}{40}$ $N_2 = 568.75 \ N$

Plugging the value of $N_2 = 568.75 \ N$ in equation (2)

$P = \frac{70(568.75)-22750}{100} \\ \\ P = \frac{39812.5-22750}{100} \\ \\ P = \frac{17062.5}{100}$

P =170.625 N

Thus; the value of horizontal force P is 170.625 N

b) From the second diagram attached the free body diagram; the free body diagram of the tool chest when it is tipping about point A is also shown below:

Taking the moments about point A:

$\sum M_A = 0$