Mechanics: Friction - Wheelbarrow Mass 8kg, Coeff. 0.6, Force 50N, Angle 30°

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The discussion focuses on the mechanics of a gardener pulling a wheelbarrow with a mass of 8kg and a coefficient of friction of 0.6. The gardener applies a force of 50N at a 30-degree angle, resulting in an acceleration of 1.29 m/s². When the wheelbarrow is loaded with an additional 20kg of soil, the total mass becomes 28kg. The analysis reveals that the horizontal component of the applied force is insufficient to overcome the increased friction force, resulting in a negative acceleration of -3.92 m/s², indicating that the wheelbarrow does not move.

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Shah 72
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A gardener is pulling a wheelbarrow of mass 8kg from rest along rough horizontal ground. The coefficient of friction between the wheelbarrow and the ground is 0.6. The gardener provides a force of 50N at an angle of 30 degree above the horizontal as shown in the diagram.
a) Find the acceleration of the wheelbarrow
I got the ans for this a = 1.29m/s^2
b) what happens when the wheelbarrow had 20kg of soil in it and the gardener exerts the same force at the same angle?
I don't understand how to solve this.
 
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Shah 72 said:
A gardener is pulling a wheelbarrow of mass 8kg from rest along rough horizontal ground. The coefficient of friction between the wheelbarrow and the ground is 0.6. The gardener provides a force of 50N at an angle of 30 degree above the horizontal as shown in the diagram.
a) Find the acceleration of the wheelbarrow
I got the ans for this a = 1.29m/s^2
b) what happens when the wheelbarrow had 20kg of soil in it and the gardener exerts the same force at the same angle?
I don't understand how to solve this.
 
mass of the wheelbarrow is now 8+20 = 28 kg

normal force increases $\implies$ friction force will increase

compare the updated friction force to the horizontal component of the applied 50N force
 
skeeter said:
mass of the wheelbarrow is now 8+20 = 28 kg

normal force increases $\implies$ friction force will increase

compare the updated friction force to the horizontal component of the applied 50N force
Thank you. So the new contact force =255N and friction is 153N. By using F=m×a, 50cos30-153=28a, I get a=-3.92m/s^2. So since the acceleration is negative the wheel barrow doesn't move. Is this reasoning correct?
 
There is no acceleration ... the horizontal component of the applied force is not great enough to overcome the maximum force of friction.
 
skeeter said:
There is no acceleration ... the horizontal component of the applied force is not great enough to overcome the maximum force of friction.
Thank you !
 

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