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

[omal3rab]

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

 

[erobz]

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]

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.

 

[erobz]

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.

 

[haruspex]

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.

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.