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](https://tex.z-dn.net/?f=F_f%3D%20%5Cmu%20N_2)
where :
friction force
= coefficient of friction
= normal friction
Given that:
= 0.3
0.3 ![N_2](https://tex.z-dn.net/?f=N_2)
Using the equation of equilibrium along horizontal direction.
![\sum f_x = 0](https://tex.z-dn.net/?f=%5Csum%20f_x%20%3D%200)
P -
0
P = 0.3
----- Equation (1)
To determine the moment about point B ; we have the expression
![\sum M_B = 0](https://tex.z-dn.net/?f=%5Csum%20M_B%20%20%3D%200)
0 = ![N_2*70-W*35-P*100](https://tex.z-dn.net/?f=N_2%2A70-W%2A35-P%2A100)
where;
P = horizontal force
= normal force at support A
W = self- weight of tool chest
Replacing W = 650 N
0 = ![N_2*70-650*35-100*P](https://tex.z-dn.net/?f=N_2%2A70-650%2A35-100%2AP)
![P = \frac{70 N_2-22750}{100} ----- equation (2)](https://tex.z-dn.net/?f=P%20%3D%20%5Cfrac%7B70%20N_2-22750%7D%7B100%7D%20-----%20equation%20%282%29)
Replacing
for P in equation (1)
![\frac{70N_2 -22750}{100} =0.3 N_2](https://tex.z-dn.net/?f=%5Cfrac%7B70N_2%20-22750%7D%7B100%7D%20%3D0.3%20N_2)
![N_2 = \frac{22750}{40}](https://tex.z-dn.net/?f=N_2%20%3D%20%5Cfrac%7B22750%7D%7B40%7D)
![N_2 = 568.75 \ N](https://tex.z-dn.net/?f=N_2%20%3D%20568.75%20%5C%20N)
Plugging the value of
in equation (2)
![P = \frac{70(568.75)-22750}{100} \\ \\ P = \frac{39812.5-22750}{100} \\ \\ P = \frac{17062.5}{100}](https://tex.z-dn.net/?f=P%20%3D%20%5Cfrac%7B70%28568.75%29-22750%7D%7B100%7D%20%5C%5C%20%5C%5C%20P%20%3D%20%5Cfrac%7B39812.5-22750%7D%7B100%7D%20%20%5C%5C%20%5C%5C%20P%20%3D%20%5Cfrac%7B17062.5%7D%7B100%7D)
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](https://tex.z-dn.net/?f=%5Csum%20M_A%20%3D%200)
-(P × 100)+ (W×35) = 0
P = ![\frac{W*35}{100}](https://tex.z-dn.net/?f=%5Cfrac%7BW%2A35%7D%7B100%7D)
Replacing 650 N for W
![P = \frac{650*35}{100}](https://tex.z-dn.net/?f=P%20%3D%20%5Cfrac%7B650%2A35%7D%7B100%7D)
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.