A block of mass

Hint and answer

In the previous problem use

Hint and answer

A conveyor is dumping sand onto a cone shaped pile. Given that the coefficient of static friction between the sand grains is

Hint and answer

A uniform ladder of length

Hint and answer

Two boards are bolted together with two bolts, as shown. The squeeze force between the boards is 500 lbs. If the shear strength of each bolt is 5000 lbs and the coefficient of static friction between the boards is

Hint and answer

A 50 kg crate is being pushed on a horizontal floor at constant velocity. Given that the coefficient of kinetic friction between crate and floor is

In the previous problem we are given that the coefficient of static friction between crate and floor is

Two children throw a rope over a tree branch and hang off each end. The children have a mass of 40 kg and 50 kg. What is the minimum coefficient of static friction between rope and tree branch so that the rope doesn't slip? To solve this consider the general equation

Hint and answer

The hints and answers for these friction problems will be given next.

The minimum force required to prevent slipping is the minimum force that will prevent the block from sliding down the incline. It is

Answer:

The maximum force pushing down the incline is 10

Answer:

The maximum sliding angle is

Answer:

This is a good static equilibrium problem. It is particularly interesting because almost everyone has stood on a ladder before, but little thought is usually given to the minimum angle to avoid slipping. It is something you just sense intuitively.

The minimum angle

Use the following sign convention: The upward and rightward direction is positive. The downward and leftward direction is negative. Counterclockwise rotation is positive. Clockwise rotation is negative.

Apply the condition of rotational equilibrium. Take the sum of the moments about the base of the ladder. This gives us:

Apply the condition of horizontal equilibrium:

Apply the condition of vertical equilibrium: -

The maximum allowable friction force at the wall is:

The maximum allowable friction force at the ground is:

Combine equations (1)-(5) to obtain an equation for

tan

Answer:

To pull the boards apart the friction force between the boards, plus the shear strength of the bolts, must be exceeded. Therefore the maximum pull force must be below the force needed to do this. Hence,

Answer: 10250 lbs

In the equation

Answer:

1.

2.

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