# Calculating Net Force on a Wheel: Magnitude and Direction | 0.350 m Radius

• omal3rab
omal3rab
Thread moved from the technical forums to the schoolwork forums
Homework Statement
Three forces are applied to a wheel of radius 0.350 m, as shown in the figure. One force is perpendicular to the rim, one is tangent to it, and the other one makes a 40.0 degree angle with the radius, and a 10 degree angle with the horizontal.

a) What is the magnitude of the net force on the wheel? [3 marks]
Relevant Equations
Fnet = sqrt(Fnetx^2 + Fnety^2)
tan(theta)= Fnety/Fnetx

Three forces are applied to a wheel of radius 0.350 m, as shown in the figure. One force is perpendicular to the rim, one is tangent to it, and the other one makes a 40.0 degree angle with the radius, and a 10 degree angle with the horizontal.

a) What is the magnitude of the net force on the wheel? [3 marks]

I am having trouble with Part a) of this problem, I got Parts b) and c) correct, but theres no answer available for a). I don't know whether I should relate Fnet to 𝜏net. I feel like I shouldn't, since the forces are not all being applied to the same point. I got an answer of Fnet = 14.4 N [-3.44°], is this correct? I made my coordinate system positive going up and to the right, made Fnetx and Fnety statements, then related them by Pythagorean Theorem.

Fnetx: 14.6cos(10) = 14.3782 N
Fnety: 8.5 + 14.6sin(10) - 11.9 = -0.86 N
theta = arctan(0.86/14.3782) = 3.44° (below horizontal, hence the negative sign above)
Fnet: sqrt((14.3782)^2 +(-0.86)^2)) = 14.404 N

Last edited:
omal3rab said:
View attachment 335410I am having trouble with Part a) of this problem, I got Parts b) and c) correct, but theres no answer available for a). I don't know whether I should relate Fnet to 𝜏net. I feel like I shouldn't, since the forces are not all being applied to the same point. I got an answer of Fnet = 14.4 N [-3.44°], is this correct? I made my coordinate system positive going up and to the right, made Fnetx and Fnety statements, then related them by Pythagorean Theorem.

Fnetx: 14.6cos(10) = 14.3782 N
Fnety: 8.5 + 14.6sin(10) - 11.9 = -0.86 N
theta = arctan(0.86/14.3782) = 3.44° (below horizontal, hence the negative sign above)
Fnet: sqrt((14.3782)^2 +(-0.86)^2)) = 14.404 N
Looks ok to me.

omal3rab
erobz said:
Looks ok to me.
Thank you! Does this mean I am correct in my assumption that I can't relate Fnet to Torque net since the forces are not all being applied to the same point? I'm still confused on what's the point of this question, since it doesn't help me solve the next part. I could be overthinking this, but it just seems odd.

omal3rab said:
Thank you! Does this mean I am correct in my assumption that I can't relate Fnet to Torque net since the forces are not all being applied to the same point? I'm still confused on what's the point of this question, since it doesn't help me solve the next part. I could be overthinking this, but it just seems odd.
It's an exercise in vector addition, I wouldn't over think it. The net torque is a different quantity from the net force.

omal3rab said:
Does this mean I am correct in my assumption that I can't relate Fnet to Torque net since the forces are not all being applied to the same point?
Yes.
omal3rab said:
what's the point of this question, since it doesn't help me solve the next part
The point is either to check you understand the difference or perhaps to drive home that they are different.

berkeman

## What is the formula to calculate the net force on a wheel?

The net force on a wheel can be calculated using the formula $$F_{\text{net}} = m \cdot a$$, where $$m$$ is the mass of the wheel and $$a$$ is the linear acceleration. For rotational motion, you might also use torque ($$\tau$$) and the radius ($$r$$) with the formula $$\tau = r \cdot F_{\text{net}}$$.

## How do you determine the direction of the net force on a wheel?

The direction of the net force on a wheel depends on the forces acting on it. If the wheel is accelerating linearly, the net force direction will be in the direction of the acceleration. For rotational motion, the direction of the net force is tangential to the wheel's edge in the direction of the applied torque.

## What role does the radius of the wheel play in calculating net force?

The radius of the wheel is crucial when calculating the torque, which is the product of the radius and the net force ($$\tau = r \cdot F_{\text{net}}$$). A larger radius means that a given force will produce a greater torque, influencing the rotational acceleration of the wheel.

## How do you account for friction when calculating the net force on a wheel?

Friction can be accounted for by including it as one of the forces in the net force calculation. The frictional force usually acts opposite to the direction of motion and can be calculated using $$F_{\text{friction}} = \mu \cdot N$$, where $$\mu$$ is the coefficient of friction and $$N$$ is the normal force. This frictional force should be subtracted from the driving force to get the net force.

## How can you calculate the linear acceleration of the wheel from the net force?

The linear acceleration of the wheel can be calculated using Newton's second law, $$a = \frac{F_{\text{net}}}{m}$$, where $$F_{\text{net}}$$ is the net force acting on the wheel and $$m$$ is the mass of the wheel. For rotational motion, you might use the relationship $$\alpha = \frac{\tau}{I}$$, where $$\alpha$$ is the angular acceleration and $$I$$ is the moment of inertia.

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