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Chapter 4: Problem 61

A wedge of mass \(m=36.1 \mathrm{~kg}\) is located on a plane that is inclined by an angle \(\theta=21.3^{\circ}\) with respect to the horizontal. A force \(F=302.3 \mathrm{~N}\) in the horizontal direction pushes on the wedge, as shown in the figure. The coefficient of kinetic friction between the wedge and the plane is 0.159 What is the acceleration of the wedge along the plane?

Short Answer

Expert verified
Answer: The acceleration of the wedge along the inclined plane is approximately 2.95 m/s².

Step by step solution

01

Diagram and Forces

Draw a diagram of the wedge on the inclined plane, with the horizontal force F acting to the right. Label the angle of inclination (θ), and indicate the weight of the wedge (mg) acting vertically downward.
02

Find the weight components

The weight of the wedge can be broken down into components parallel and perpendicular to the inclined plane: - Parallel component: \(mg \sin \theta\) - Perpendicular component: \(mg \cos \theta\)
03

Calculate the friction force

The friction force acting on the wedge is given by: - \(F_f = \mu F_n\) Where F_f is the friction force and F_n is the normal force. In this case, the normal force is equal to the perpendicular component of the weight (mg cos(θ)): \(F_f = \mu (mg \cos \theta)\)
04

Resolve the horizontal force

The horizontal force F can also be broken down into components parallel and perpendicular to the inclined plane: - Parallel component: \(F \cos \theta\) - Perpendicular component: \(F \sin \theta\)
05

Apply Newton's Second Law

Now we apply Newton's second law along the parallel direction, taking into account the horizontal force component, weight component, and the friction force: - \(ma_{\parallel} = F_{\parallel} - mg_{\parallel} - F_f\) Substitute the known values and the expressions derived in Steps 2, 3, and 4: \(m a_{\parallel} = (F \cos \theta) - (mg \sin \theta) - (\mu (mg \cos \theta))\)
06

Solve for acceleration

Now, solve for the acceleration along the plane (a_𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙). \(a_{\parallel} = \frac{(F \cos \theta) - (mg \sin \theta) - (\mu (mg \cos \theta))}{m}\) Plug in the given values of m, θ, F, and μ: \(a_{\parallel} = \frac{(302.3 \mathrm{~N} \cos 21.3^{\circ}) - (36.1 \mathrm{~kg} \cdot 9.81 \mathrm{~m/s^2} \sin 21.3^{\circ}) - (0.159 (36.1 \mathrm{~kg} \cdot 9.81 \mathrm{~m/s^2} \cos 21.3^{\circ}))}{36.1 \mathrm{~kg}}\) After calculating the expression, we find: \(a_{\parallel} \approx 2.95 \mathrm{~m/s^2}\) The acceleration of the wedge along the inclined plane is approximately \(2.95 \mathrm{~m/s^2}\).

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