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Chapter 4: Problem 112

You want to push a \(65-\mathrm{kg}\) box up a \(25^{\circ}\) ramp. The coefficient of kinetic friction between the ramp and the box is \(0.30 .\) With what magnitude force parallel to the ramp should you push on the box so that it moves up the ramp at a constant speed?

Short Answer

Expert verified
Answer: To push the box up the ramp at a constant speed, you should apply a force of approximately 249.5 N parallel to the ramp.

Step by step solution

01

Identify the forces acting on the box

First, let's decompose the gravitational force acting on the box into two components: one parallel to the ramp and another perpendicular to it. The force of gravity is given by \(F_g = mg\), where m is the mass of the box, and g is the acceleration due to gravity. The component parallel to the ramp will be \(F_{g\parallel} = mg\sin(25^{\circ})\) and the component perpendicular to the ramp will be \(F_{g\perp} = mg\cos(25^{\circ})\). Also, note that the force of friction is given by \(F_f = \mu F_N\), where \(F_N\) is the normal force acting on the box, and \(\mu\) is the coefficient of kinetic friction.
02

Apply Newton's second law of motion

Since the box moves up the ramp at a constant speed, the net force along the ramp is zero. Therefore, the applied force (\(F_{app}\)) is equal to the sum of the component of the gravitational force parallel to the ramp and the frictional force: \(F_{app} = F_{g\parallel} + F_f\).
03

Calculate the normal force

The normal force acting on the box is equal to the component of the gravitational force perpendicular to the ramp: \(F_N = F_{g\perp} = mg\cos(25^{\circ})\).
04

Calculate the frictional force

Now that we know the normal force, we can find the force of friction: \(F_f = \mu F_N = \mu mg\cos(25^{\circ})\).
05

Calculate the applied force

Finally, plug the values for \(F_{g\parallel}\) and \(F_f\) into the equation for the applied force: \(F_{app} = mg\sin(25^{\circ}) + \mu mg\cos(25^{\circ})\). Using the given values for the mass (\(m\)), the angle of the ramp (\(25^{\circ}\)), and the coefficient of kinetic friction (\(\mu\)), we can compute the applied force: \(F_{app} = (65\text{ kg})(9.81\frac{\text{m}}{\text{s}^2})\sin(25^{\circ}) + 0.30 (65\text{ kg})(9.81\frac{\text{m}}{\text{s}^2})\cos(25^{\circ})\) Upon calculation, we find that \(F_{app} \approx 249.5\text{ N}\). Therefore, you should push the box with a force of approximately \(249.5\text{ N}\) parallel to the ramp in order for it to move up the ramp at a constant speed.

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