8. A person wants to push a 110-kg box across the floor. The coefficient of fric

Question

8. A person wants to push a 110-kg box across the floor. The coefficient of friction between the box and the floor is 0.3. a. What is the normal force experienced by the box at rest in [N]? b. If the person pushes horizontally on the box, how much force must be applied to move the box in [N]? C. If the person pushes against the box with 500N of force but at an angle of 20% relative to the horizontal, what is the normal force experienced by the box in [N]? d. What is the horizontal acceleration of the box when the 500N of force at an angle of 20% relative to the horizontal is applied, in [m/s]?

Explanation / Answer

(a) m = 110 kg, mu = 0.3

Normal force experienced by the box at rest R = mg = 110*9.8 = 1078 N

(b) If the person pushes the box horizontally, the requite force required = frictional force acts on the box

= mu*m*g = 0.3*1078 = 323.4 N

(c) Here, F = 500 N

Horizontal component of the force, Fx = F*cos20 = 470 N

Vertical component of the force, Fy = F*sin20 = 171 N

So, normal force experienced by the box = mg+ Fy = 1078 + 171 = 1249 N

(d) Horizontal acceleration = [Fx - mu(mg+Fy)] / m = [470 - 0.3*1249] / 110 = 0.87 m/s^2

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