Solution to problem 2.6.5 from the collection of Kepe O.E.

2.6.5 In the problem, we are given a roller 1 to which a load 2 is attached using an inextensible thread. It is necessary to determine the greatest weight of the load at which roller 1 weighing 3.2 kilonewtons will remain at rest. The rolling friction coefficient between the roller and the surface is ? = 0.004, and the radius of the skating rink R = 32.4 centimeters.

Solution: Since the roller is at rest, the frictional force between the roller and the surface is equal to the tension in the thread. The friction force is equal to the product of the rolling friction coefficient and the normal pressure force, which is equal to the weight of the roller. Thus, the tension in the thread is equal to the weight of the load plus the weight of the roller multiplied by the coefficient of friction.

You can write the equilibrium equation in the form: Fn - Ftr = 0, where Fn is the tension force of the thread, Ftr is the friction force.

Let's express the friction force: Ftr = ? * Fn, where? - rolling friction coefficient.

Let's substitute the value of the friction force into the equilibrium equation: Fн - ? * Fн = 0.

Let us express the tension force of the thread: Fн = mg * g, where mg is the mass of the load, g is the acceleration of free fall.

Let's substitute the value of the tension force of the thread into the equilibrium equation: mg * g - ? * mg * g * R = 0.

Let us express the mass of the load: mg = Fк / g, where Fк is the largest weight of the load at which the roller will remain at rest.

Let's substitute the value of the load mass into the equilibrium equation: Fк = ? * mg * g * R = ? * Fк * R * g / ?.

Solving this equation, we obtain: Fk = (3.2 * 1000 * 9.81) / (0.004 * 0.324 * 2) = 399.5 N, which is rounded to 40.0 kilonewtons.

Thus, the maximum weight of the load that can be suspended from roller 1 under these conditions is 40.0 kilonewtons.

Solution to problem 2.6.5 from the collection of Kepe O.?.

Solution to problem 2.6.5 from the collection of Kepe O.?. is a digital product intended for students, students and anyone interested in physics. This solution explains in detail how to find the largest weight of a load that can be suspended from roller 1 using an inextensible thread under given conditions.

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Solution to problem 2.6.5 from the collection of Kepe O.?. consists in calculating the maximum weight of the load that can be suspended from roller 1 so that it remains at rest.

To solve the problem it is necessary to use the laws of mechanics. According to the condition, roller 1 is subject to the force of gravity of the load, equal to the weight of the load, and the rolling friction force, which is proportional to the normal pressure force and the rolling friction coefficient. In the absence of acceleration of the roller, the sum of all forces acting on it is equal to zero.

Using the formula for calculating the rolling friction force and applying the law of conservation of energy, we can obtain an equation in which the unknown is the weight of the load. By solving this equation, we can determine the maximum weight of the load that can be suspended from roller 1 so that it remains at rest.

The result of solving the problem is the number 40.0, which is the maximum weight of the load that can be suspended from roller 1 so that it remains at rest under the given conditions.

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