# Simple force with friction question

## Homework Statement

The person in the http://images3a.snapfish.com/232323232fp733:2>nu=52::>379>256>WSNRCG=335:3:<794347nu0mrj" 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fp733;6>nu=52::>379>256>WSNRCG=335:453:76347nu0mrj" 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.

## The Attempt at a Solution

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

$$F_{net}=0$$

$$F_{\mu}=\mu*F_{n}$$

$$F_{net_{y}} = mg-F_{p}sin(\theta) - \mu F_{n}=0$$

$$F_{net_{x}}= F_{p}cos(\theta) - F_{n} = 0$$

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

$$-\mu F_{n}=-mg+F_{p}sin(\theta)$$

$$\mu F_{n}=mg-F_{P}sin(\theta)$$

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

$$F_{p}sin(\theta)=F_{n}$$

$$F_{p}= \frac {F_{n}}{cos(\theta)}$$

$$\mu F_{n}=mg-(\frac{F_{n}}{cos(\theta)})sin(\theta)$$

$$\mu F_{n}=mg-F_{n}tan(\theta)$$

$$\mu F_{n}+F_{n}tan(\theta)=mg$$

$$F_{n}(\mu+tan(\theta))=mg$$

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

$$F_{P}= \frac {F_{n}}{cos(\theta)} = \frac {850}{cos(30)} \approx 981.5 N$$

$$F_{\mu} = \mu * F_{n} = .6 * 981.5N \approx 510N$$

is that the correct way to think about the problem?

thanks

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SammyS
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Homework Helper
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## Homework Statement

The person in the http://images3a.snapfish.com/232323232fp733:2>nu=52::>379>256>WSNRCG=335:3:<794347nu0mrj" 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fp733;6>nu=52::>379>256>WSNRCG=335:453:76347nu0mrj" 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.

## The Attempt at a Solution

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

$$F_{net}=0$$

$$F_{\mu}=\mu*F_{n}$$

$$F_{net_{y}} = mg-F_{p}sin(\theta) - \mu F_{n}=0$$

$$F_{net_{x}}= F_{p}cos(\theta) - F_{n} = 0$$

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

$$-\mu F_{n}=-mg+F_{p}sin(\theta)$$

$$\mu F_{n}=mg-F_{P}sin(\theta)$$

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

$$F_{p}sin(\theta)=F_{n}$$  $$\text{This should be } F_{p}\cos(\theta)=F_{n}\,.$$

$$F_{p}= \frac {F_{n}}{cos(\theta)}$$

$$\mu F_{n}=mg-(\frac{F_{n}}{cos(\theta)})sin(\theta)$$

$$\mu F_{n}=mg-F_{n}tan(\theta)$$

$$\mu F_{n}+F_{n}tan(\theta)=mg$$

$$F_{n}(\mu+tan(\theta))=mg$$

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

$$F_{P}= \frac {F_{n}}{cos(\theta)} = \frac {850}{cos(30)} \approx 981.5 N$$

$$F_{\mu} = \mu * F_{n} = .6 * 981.5N \approx 510N$$

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.

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lol strange that i changed it from sin to cos in the next line but cool. The weight is 1000N down. Thanks!