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

• MHB
• Shah 72
In summary, a gardener pulling a wheelbarrow of mass 8kg with a coefficient of friction of 0.6 along rough horizontal ground with a force of 50N at an angle of 30 degrees above the horizontal will result in an acceleration of 1.29m/s^2. If the wheelbarrow has an additional 20kg of soil and the same force is applied, the normal force and friction force will increase, resulting in no acceleration as the horizontal component of the applied force is not enough to overcome the maximum friction force.
Shah 72
MHB
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.

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 !

## 1. What is the formula for calculating friction force in this scenario?

The formula for calculating friction force is F = μN, where F is the friction force, μ is the coefficient of friction, and N is the normal force.

## 2. How do you determine the normal force in this situation?

The normal force can be determined by multiplying the mass of the object (in this case, 8kg) by the acceleration due to gravity (9.8m/s²), since the object is on a flat surface and not accelerating vertically. Therefore, the normal force in this scenario would be 78.4N.

## 3. What is the direction of the friction force in this case?

The direction of the friction force is always opposite to the direction of motion or attempted motion. In this scenario, since the wheelbarrow is being pushed forward, the friction force would act in the opposite direction, pushing backwards.

## 4. How does the angle of the incline affect the friction force?

The angle of the incline affects the normal force, which in turn affects the friction force. As the angle increases, the normal force decreases, resulting in a decrease in the friction force. In this case, the angle of 30° would result in a normal force of 78.4N * cos(30°) = 67.9N and a friction force of 0.6 * 67.9N = 40.7N.

## 5. How does the coefficient of friction affect the force needed to move the wheelbarrow?

The coefficient of friction is a measure of the roughness or smoothness of the surfaces in contact. A higher coefficient of friction means there is more resistance to motion, so a greater force would be needed to move the wheelbarrow. In this scenario, a coefficient of 0.6 means that 60% of the normal force is required to overcome the friction, so the force needed to move the wheelbarrow would be 60% of 78.4N, or 47.0N.

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