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Chapter 4: Q77. (page 106)

In the design of a supermarket, there are to be several ramps connecting different parts of the store. Customers will have to push grocery carts up the ramps and it is desirable that this not be too difficult. The engineer has done a survey and found that almost no one complaints if the force required is no more than 18 N. Ignoring friction, at what maximum angle胃should the ramps be built, assuming a full 25-kg cart?

Short Answer

Expert verified

The maximum angle 胃can be 4.68掳.

Step by step solution

01

Step 1. Understanding equilibrium

When the net force acting on a body is zero, the body is said to be in equilibrium.

02

Step 2. Given data and assumptions

The mass of the full cart,M=25kg

Let the angle of the ramp be represented by angle 胃, and the force applied by the customer be represented by F.

03

Step 3. Drawing the FBD of the cart

The FBD of the cart placed on the ramp can be drawn as:

Here, N is the normal reaction by the ramp on the cart.

04

Step 4. Finding out the maximum value of angle

Applying Newton鈥檚 second law along the incline, you get:

F=Mgsin胃

As the value of F should be less than or equal to 18 N, so

Mgsin胃鈮�18.

Substituting the values in the above equation, you get:

25脳9.8脳sin胃鈮�18

From the above equation,

sin胃鈮�0.073

And,

胃鈮�4.68掳

Thus, the maximum value of the angle is 4.68掳.

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Most popular questions from this chapter

Question: (II) Uphill escape ramps are sometimes provided at the side of steep downhill highways for trucks with overheated brakes. For a simple 11掳 upward ramp, what minimum length would be needed for a runaway truck traveling at 140 km/h. Note the large size of your calculated length. (If sand is used for the bed of the ramp, its length can be reduced by a factor of about 2.)

A truck is traveling horizontally to the right (Fig.4鈥�38). When the truck starts to slow down, the crate on the (frictionless) truck bed starts to slide. In what direction could the net force be on the crate?(a) No direction. The net force is zero.(b) Straight down (because of gravity).(c) Straight up (the normal force).(d) Horizontal and to the right.(e) Horizontal and to the left.

Two crates, of mass 65 kg and 125 kg, are in contact and at rest on a horizontal surface (Fig. 4鈥�57). A 650-N force is exerted on the 65-kg crate. If the coefficient of kinetic friction is 0.18, calculate (a) the acceleration of the system, and (b) the force that each crate exerts on the other (c) Repeat with the crates reversed.

FIGURE 4-57Problem 49.

Matt, in the foreground of Fig. 4-39, is able to move the large truck because

(a) he is stronger than the truck.

(b) he is heavier in some respect than the truck.

(c) he exerts a greater force on the truck than the truck exerts on him.

(d) the ground exerts greater friction on Matt than it does on the truck.

(e) the truck offers no resistance because its brakes are off.

A heavy crate rests on the bed of a flatbed truck. When the truck accelerates, the crate stays fixed on the truck, so it, too, accelerates. What force causes the crate to accelerate?
See all solutions

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