A 120-lb cabinet is mounted on casters that can be locked to prevent their rotation. The coefficient of static friction between the floor and each caster is 0.30. Assuming that the casters at both A and B are locked, detemine
a. The force P required to move the cabinet to the right,
b. The largest allowable value of h if the cabinet is not to tip over.

a)36 lb

b) 40 in

Explanation:

Given that;

Weight of the cabinet (W) = 120 lb

co-efficient of friction (μ) = 0.30

The force P required to move the cabinet to the right can be determined by the following:

Equilibrium Equation along the horizontal direction to the right is equal to zero.

Fricitonal force (F) = μN

where μ = co-efficient of friction & N = normal reaction

Now;

+ - W = 0

W =   +

W = weight of the cabinet

So for locked caster A and B , the normal reaction are and respectively.

Since;

where:

= frictional force for caster A;     &

= frictional force for caster B

So, that implies that;

P -  μ -  μ = 0

P = μ +  μ

P = μ

we can as well say taht:

P = μW       since W =

P = 0.3 × 120 lb

P = 36 lb

We can now say that, the  force P required to move the cabinet to the right = 36 lb

b)

Determine; The largest allowable value of h if the cabinet is not to tip over.

So if the cabinet tip over point B where it is being locked, point A definitely loses contact with the ground.

Then there exist no reaction exerted by the ground surface at point A.

So, we have to take a look at the moment equilibrium equation about point B which can be represented as:

=     (since the difference between = 24 in

h =

If W = 120lb and P = 36 lb

then h =

h = 40 in

∴ The largest allowable value of h if the cabinet is not to tip over = 40 in

answer:

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Explanation: