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

Click For Summary
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, increasing the normal force and friction. The horizontal component of the applied force is insufficient to overcome the maximum friction force, leading to a calculated negative acceleration of -3.92 m/s², indicating that the wheelbarrow does not move. This analysis confirms that the applied force is inadequate to initiate movement under the increased load.
Shah 72
MHB
Messages
274
Reaction score
0
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.
 
Mathematics news on Phys.org
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 !
 

Similar threads

MHB Find Magnitude of Total Contact Force on 20kg Box in 3 Cases
Replies
4
Views
2K
MHB Solving Mechanics Problem: Friction, 2 Boys, Toy Acceleration
Replies
4
Views
1K
MHB Solving Mechanics: Friction of Sledge on Slope
Replies
9
Views
1K
MHB Boy Exerts Force on 6kg Chair to Prevent Slipping
Replies
2
Views
1K
MHB Friction Force: 1.3 kg Book on 16° Plank - Find the Answer!
Replies
2
Views
1K
MHB Mass 50kg Box Slows to Halt on Friction: 14.5m Distance
Replies
4
Views
1K
MHB Mechanics: Friction - 200kg Bag Winched up 10m Slope at 6°
Replies
8
Views
1K
MHB Find Coefficient of Friction for Toy Car Rolling Down Slope
Replies
6
Views
2K
MHB Solving Friction Problem: Mass 5kg Trolley, 25° Slope, Coefficient 0.4
Replies
2
Views
1K
What Are the Forces Needed to Hold a Wheelbarrow at a 30∘ Angle?
Replies
1
Views
4K