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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:

$\sum M_A = 0$

-(P × 100)+ (W×35) = 0

P = $\frac{W*35}{100}$

Replacing 650 N for W

$P = \frac{650*35}{100}$

P = 227.5 N

Thus; the value of horizontal force P, when the tool chest tipping about point A is 227.5 N

We conclude that the motion will be impending for the lowest value when P = 170.625 N and when P= 227.5 N

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

5 0

3 years ago

What is the lehr and what purpose does it serve?

levacccp [35]

Explanation:

Step1

Lehr is the long open or closed insulated space for glass manufacturing. Lehr must be large enough to keep the cooling of glass uniform. The function of Lehr is the same as an annealing process in metallurgy.

Step2

Lehr decrease the cooling and temperature variation in glass production. Uneven temperature creates the internal stress in the glass. Lehr reduces the internal stress in the glass product. So, the main purpose of the Lehr is to reduce the internal stress and keep the cooling uniform.

5 0

3 years ago

what would the current through the Load curve look like when the alternating current source S has the following shape Draw the s

IgorLugansk [536]

Answer:

Following are the response to the given question:

Explanation:

In the given question, A curve one would be to "reject" them. Within this way, people are directly non-confrontationally off from their amorous interests or advances. The curve looks much like a current through the unidirectional load, and it has only a good maximum value. It seems like the total rating of the complete wave rectifier that is illustrated below, the present vas temporal curve please find it.