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. The surface has a coefficient of friction of 0.3. Determine the value of the horizontal force P necessary to cause motion of the chest to the right, and determine if the motion is sliding or tipping.

Respuesta :

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

[tex]F_f= \mu N_2[/tex]

where :

[tex]F_f =[/tex] friction force

[tex]\mu[/tex] = coefficient of friction

[tex]N_2[/tex] = normal friction

Given that:

[tex]\mu[/tex] = 0.3

[tex]F_f =[/tex] 0.3 [tex]N_2[/tex]

Using the equation of equilibrium along horizontal direction.

[tex]\sum f_x = 0[/tex]

P - [tex]F_f =[/tex] 0

P = 0.3 [tex]N_2[/tex]   ----- Equation (1)

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

[tex]\sum M_B = 0[/tex]

0 = [tex]N_2*70-W*35-P*100[/tex]

where;

P = horizontal force

[tex]N_2[/tex] = normal force at support A

W = self- weight of tool chest

Replacing W = 650 N

0 = [tex]N_2*70-650*35-100*P[/tex]

[tex]P = \frac{70 N_2-22750}{100} ----- equation (2)[/tex]

Replacing  [tex]\frac{70 N_2-22750}{100}[/tex]  for P in equation (1)

[tex]\frac{70N_2 -22750}{100} =0.3 N_2[/tex]

[tex]N_2 = \frac{22750}{40}[/tex][tex]N_2 = 568.75 \ N[/tex]

Plugging the value of [tex]N_2 = 568.75 \ N[/tex] in equation (2)

[tex]P = \frac{70(568.75)-22750}{100} \\ \\ P = \frac{39812.5-22750}{100} \\ \\ P = \frac{17062.5}{100}[/tex]

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:

[tex]\sum M_A = 0[/tex]

-(P × 100)+ (W×35) = 0

P = [tex]\frac{W*35}{100}[/tex]

Replacing 650 N  for W

[tex]P = \frac{650*35}{100}[/tex]

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.

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