Simple force with friction question

In summary, the conversation discusses a problem where a person is exerting the minimum force needed to keep a 100 kg block from falling, while pushing at an angle of 30 degrees with respect to the horizontal. The coefficient of friction is 0.6 and the goal is to determine the magnitudes of all the forces involved. The solution involves setting the net force to 0 and using trigonometry to solve for the normal force and friction force. The final answer is approximately 850 N for the normal force, 981.5 N for the applied force, and 510 N for the friction force. The weight of the object, 1000 N, is also mentioned.
  • #1
Asphyxiated
264
0

Homework Statement



The person in the http://images3a.snapfish.com/232323232%7Ffp733%3A2%3Enu%3D52%3A%3A%3E379%3E256%3EWSNRCG%3D335%3A3%3A%3C794347nu0mrj" is exerting the minimum force needed to keep the 100 kg block from falling down the wall. She is pushing at an angle of 30 degrees with respect to the horizontal. The coefficient of friction is 0.6. Determine the magnitudes of all the forces in the problem.

I made a http://images3a.snapfish.com/232323232%7Ffp733%3B6%3Enu%3D52%3A%3A%3E379%3E256%3EWSNRCG%3D335%3A453%3A76347nu0mrj" where Fp is the force the person is putting on the box, Fpcos and Fpsin are the respective x and y components of Fp. Ffriction is the solid black line going up and Fn is the normal force from the x component of Fp and m*g is the force due to gravity on the object.

Homework Equations


The Attempt at a Solution



The object is not in motion so all Fnets should be 0.

[tex] F_{net}=0 [/tex]

[tex] F_{\mu}=\mu*F_{n} [/tex]

[tex] F_{net_{y}} = mg-F_{p}sin(\theta) - \mu F_{n}=0 [/tex]

[tex] F_{net_{x}}= F_{p}cos(\theta) - F_{n} = 0 [/tex]

There may be an easier way to the end from here but this is what I did:

[tex] -\mu F_{n}=-mg+F_{p}sin(\theta) [/tex]

[tex] \mu F_{n}=mg-F_{P}sin(\theta) [/tex]

now i will solve the other equation for Fp and sub it in:

[tex] F_{p}sin(\theta)=F_{n} [/tex]

[tex] F_{p}= \frac {F_{n}}{cos(\theta)} [/tex]

[tex] \mu F_{n}=mg-(\frac{F_{n}}{cos(\theta)})sin(\theta) [/tex]

[tex] \mu F_{n}=mg-F_{n}tan(\theta) [/tex]

[tex] \mu F_{n}+F_{n}tan(\theta)=mg [/tex]

[tex] F_{n}(\mu+tan(\theta))=mg [/tex]

[tex] F_{n}= \frac {mg}{\mu+tan(\theta)} = \frac {(100)(10)}{.6+tan(30)} \approx 850 N [/tex]

[tex] F_{P}= \frac {F_{n}}{cos(\theta)} = \frac {850}{cos(30)} \approx 981.5 N [/tex]

[tex] F_{\mu} = \mu * F_{n} = .6 * 981.5N \approx 510N [/tex]

is that the correct way to think about the problem?

thanks
 
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  • #2
Asphyxiated said:

Homework Statement



The person in the http://images3a.snapfish.com/232323232%7Ffp733%3A2%3Enu%3D52%3A%3A%3E379%3E256%3EWSNRCG%3D335%3A3%3A%3C794347nu0mrj" is exerting the minimum force needed to keep the 100 kg block from falling down the wall. She is pushing at an angle of 30 degrees with respect to the horizontal. The coefficient of friction is 0.6. Determine the magnitudes of all the forces in the problem.

I made a http://images3a.snapfish.com/232323232%7Ffp733%3B6%3Enu%3D52%3A%3A%3E379%3E256%3EWSNRCG%3D335%3A453%3A76347nu0mrj" where Fp is the force the person is putting on the box, Fpcos and Fpsin are the respective x and y components of Fp. Ffriction is the solid black line going up and Fn is the normal force from the x component of Fp and m*g is the force due to gravity on the object.

Homework Equations





The Attempt at a Solution



The object is not in motion so all Fnets should be 0.

[tex] F_{net}=0 [/tex]

[tex] F_{\mu}=\mu*F_{n} [/tex]

[tex] F_{net_{y}} = mg-F_{p}sin(\theta) - \mu F_{n}=0 [/tex]

[tex] F_{net_{x}}= F_{p}cos(\theta) - F_{n} = 0 [/tex]

There may be an easier way to the end from here but this is what I did:

[tex] -\mu F_{n}=-mg+F_{p}sin(\theta) [/tex]

[tex] \mu F_{n}=mg-F_{P}sin(\theta) [/tex]

now i will solve the other equation for Fp and sub it in:

[tex] F_{p}sin(\theta)=F_{n} [/tex]  [tex]\text{This should be } F_{p}\cos(\theta)=F_{n}\,.[/tex]

[tex] F_{p}= \frac {F_{n}}{cos(\theta)} [/tex]

[tex] \mu F_{n}=mg-(\frac{F_{n}}{cos(\theta)})sin(\theta) [/tex]

[tex] \mu F_{n}=mg-F_{n}tan(\theta) [/tex]

[tex] \mu F_{n}+F_{n}tan(\theta)=mg [/tex]

[tex] F_{n}(\mu+tan(\theta))=mg [/tex]

[tex] F_{n}= \frac {mg}{\mu+tan(\theta)} = \frac {(100)(10)}{.6+tan(30)} \approx 850 N [/tex]

[tex] F_{P}= \frac {F_{n}}{cos(\theta)} = \frac {850}{cos(30)} \approx 981.5 N [/tex]

[tex] F_{\mu} = \mu * F_{n} = .6 * 981.5N \approx 510N [/tex]

is that the correct way to think about the problem?

thanks
Method looks good.

A little typo as indicated in the Quote above.

One other force: Weight.
 
Last edited by a moderator:
  • #3
lol strange that i changed it from sin to cos in the next line but cool. The weight is 1000N down. Thanks!
 

1. What is a simple force with friction?

A simple force with friction refers to the combination of two basic forces: a force acting on an object and friction acting in the opposite direction. Friction is a force that resists the motion of an object when it is in contact with another surface.

2. How is friction calculated?

Friction is calculated by multiplying the coefficient of friction, which is a measure of how rough or smooth two surfaces are in contact, by the normal force, which is the force perpendicular to the surface that the object is resting on.

3. What factors affect the amount of friction?

The amount of friction is affected by the coefficient of friction, the normal force, and the surface area of the object in contact with the surface. Rougher surfaces, higher normal forces, and larger surface areas result in higher amounts of friction.

4. How does friction affect the motion of an object?

Friction acts in the opposite direction of the object's motion, causing it to slow down or come to a stop. It also affects the amount of force needed to move an object, as more force is required to overcome the resistance of friction.

5. How can friction be reduced?

Friction can be reduced by using lubricants, such as oil or grease, to create a smoother surface between two objects in contact. Additionally, decreasing the normal force or using smoother surfaces can also reduce friction.

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