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.

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

where :

friction force

= coefficient of friction

= normal friction

Given that:

= 0.3

0.3

Using the equation of equilibrium along horizontal direction.

P - 0

P = 0.3   ----- Equation (1)

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

0 =

where;

P = horizontal force

= normal force at support A

W = self- weight of tool chest

Replacing W = 650 N

0 =

Replacing    for P in equation (1)

Plugging the value of in equation (2)

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:

-(P × 100)+ (W×35) = 0

P =

Replacing 650 N  for W

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.

can you make it more obvious. i can't see it properly

answer:

a

explanation: