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Chapter 8: Problem 18

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, determine (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.

Short Answer

Expert verified
(a) The force required is 36 lb. (b) The largest allowable height is \( \frac{5l}{3} \), assuming the horizontal distance \( l \) between casters is known.

Step by step solution

01

Calculate the Normal Force at the Casters

The weight of the cabinet is given as 120 lb. Assuming the weight is evenly distributed between the two casters (A and B), the normal force at each caster is half of the total weight. Therefore, the normal force at each caster is \( N_A = N_B = \frac{120}{2} = 60 \text{ lb} \).
02

Determine the Static Friction Force

The frictional force at each caster is determined using the formula \( f = \mu_s N \), where \( \mu_s = 0.30 \) is the coefficient of static friction. Thus, the frictional force per caster is \( f = 0.30 \times 60 = 18 \text{ lb} \). The total frictional force to overcome is then the sum of forces at both casters, \( 2f = 36 \text{ lb} \).
03

Compute the Force P Required to Slide the Cabinet

Force \( P \) must overcome the total static friction force of 36 lb. Therefore, \( P = 36 \text{ lb} \) is required to move the cabinet to the right, as it is equal to the total frictional force.
04

Analyze the Tipping Condition

The tipping condition will be met when the force creates a moment about one caster that exceeds the moment created by the cabinet's weight about the same caster. Assume tipping point about caster A; thus, \( P \cdot h = W \cdot \frac{l}{2} \), where \( W = 120 \text{ lb} \) and \( l \) is the horizontal distance between casters. The force creates a counterclockwise moment, and the weight creates a clockwise moment.
05

Calculate the Largest Allowable Height \\( h \\\)

Assume the cabinet is tipping around caster A. The equation \( P \cdot h = 120 \cdot \frac{l}{2} \) gives \( h = \frac{60l}{P} \). With \( P = 36 \text{ lb} \), \( h \leq \frac{60l}{36} = \frac{5l}{3} \). Assuming \( l \) is the distance between casters; without its value, we can express the relationship. For a specific answer, \( l \) must be given.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Normal Force
When understanding the movement of objects, normal force is a fundamental concept. It plays a crucial role in situations involving static friction. Normal force, denoted as \( N \), is the perpendicular force exerted by a surface in reaction to an object's weight.

In our example, the cabinet weighs 120 lb. When the cabinet is mounted on two casters, the weight is evenly distributed between them. Each caster supports half of the cabinet鈥檚 weight, resulting in a normal force of 60 lb for each.

This force is essential in calculating the static friction force, which is the resistance an object encounters when it attempts to slide across a surface. By using the normal force and the coefficient of static friction (\( \mu_s \)), we determine how much resistance needs to be overcome to move the cabinet.
Tipping Condition
Tipping condition is a critical concept when assessing whether or not an object might fall over. It occurs when a force causes an object to pivot beyond its base of support.

In the context of our exercise, as the cabinet is pushed, it could either slide or tip over. Tipping begins when a force creates a moment that competes with the object's weight.

To quantify the tipping condition, consider a point (e.g., caster on one side) where the cabinet might pivot. If the force applied causes a greater moment about this point than the moment from the cabinet's weight, tipping will happen. It鈥檚 crucial to balance these forces to prevent the cabinet from toppling over.
Moment of Force
Moment of force, often referred to as torque, is the rotational effect of a force applied at a distance from a pivot point. It is calculated using the formula \( M = F \times d \), where \( F \) is the force and \( d \) is the distance from the pivot point.

For the cabinet in our scenario, when applying a sideward force, it introduces a moment about the edge of one of the casters. The tipping condition equation, \( P \cdot h = W \cdot \frac{l}{2} \), involves equating the moment from the force \( P \) at height \( h \) with the moment from the weight \( W \) about the caster's pivot point.

Understanding the moment of force helps in determining the safest way to apply force without causing tipping, especially when the height at which the force is applied affects stability.

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Most popular questions from this chapter

A \(5^{\circ}\) wedge is to be forced under a 1400 -lb machine base at \(A\). Knowing that the coefficient of static friction at all surfaces is \(0.20,\) (a) determine the force \(P\) required to move the wedge, \((b)\) indicate whether the machine base will move.

Two slender rods of negligible weight are pin-connected at \(C\) and attached to blocks \(A\) and \(B\), each with a weight \(W .\) Knowing that \(\theta=80^{\circ}\) and that the coefficient of static friction between the blocks and the horizontal surface is 0.30 , determine the largest value of \(P\) for which equilibrium is maintained.

A 900-kg machine base is rolled along a concrete floor using a series of steel pipes with outside diameters of 100 mm. Knowing that the coefficient of rolling resistance is 0.5 mm between the pipes and the base and 1.25 mm between the pipes and the concrete floor, determine the magnitude of the force P required to slowly move the base along the floor.

A \(12^{\circ}\) wedge is used to spread a split ring. The coefficient of static friction between the wedge and the ring is \(0.30 .\) Knowing that a force \(P\) with a magnitude of \(120 \mathrm{N}\) was required to insert the wedge, determine the magnitude of the forces exerted on the ring by the wedge after insertion.

A rope having a weight per unit length of \(0.4 \mathrm{lb} / \mathrm{ft}\) is wound \(2 \frac{1}{2}\) times around a horizontal rod. Knowing that the coefficient of static friction between the rope and the rod is \(0.30,\) determine the minimum length \(x\) of rope that should be left hanging if a \(100-1 b\) load is to be supported.
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