# Friction Bonus Problem

1. Oct 22, 2008

### burg25

1. The problem statement, all variables and given/known data
Hey I was wondering if someone would just check over this bonus question for me.
1) A father pulls his child on a sled by pulling on rope over his shoulder. He exerts a force of 50 N at an angle of 30*. a) Using the coefficient of friction for steel on ice, calculate the acceleration of the sled if its mass plus the child's is 20 kg, assuming the sled is already in motion. b) What minimum force must the father exert to start the sled moving from rest?

2) Block B in the figure (in attached file) weighs 711 N. The Coefficient of static friction between the block and the horizontal surface is 0.25. Find the maximum weight of block A for which the system will be stationary. Assume the rope attached to the wall make a 30* angle with the horizontal.
2. Relevant equations
$$\mu$$s = 0.40 $$\mu$$k = 0.02
F=ma

3. The attempt at a solution
1) a) Fx= max Fy=may
Fappcos$$\theta$$-f = ma Fn+Fappsin$$\theta$$= Fg
Fappcos$$\theta$$- $$\mu$$kFn=ma Fn = Fg - Fappsin$$\theta$$
a = Fappcos$$\theta$$- $$\mu$$kFn/m
a = (50cos30 - (0.02(20(9.8)- 50 sin30))/20
a = 1.99m/s/s
b) Fcos$$\theta$$- $$\mu$$s (Fg-Fsin$$\theta$$) = 0
Fcos$$\theta$$- $$\mu$$sFg +$$\mu$$sFsin$$\theta$$)=0
Fcos$$\theta$$+$$\mu$$sFsin$$\theta$$)=$$\mu$$sFg
F(cos$$\theta$$+$$\mu$$ssin$$\theta$$) =$$\mu$$sFg
F =$$\mu$$sFg / (cos$$\theta$$+$$\mu$$ssin$$\theta$$
F = (0.4(20*9.8)) / (cos30 + (0.4(sin30))
F = 73.5 N
2) I was confused on this one and treated it as a hanging mass problem but here is my attempt:
Fb=ma
Fb=mg-$$\mu$$kmg
Fb= 711 - (0.25(711))
Fb=533.25N
Tbsin$$\theta$$= Tacos$$\theta$$
Ta = Tbsin$$\theta$$ / cos$$\theta$$
Ta = 533.25sin30 / cos30
Ta = 307.9N
The maximum weight for block A is 307.9N because if it were greater the string would break.

#### Attached Files:

• ###### Doc2.doc
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2. Oct 23, 2008

### alphysicist

Hi burg25,

Your work for the first problem looks okay to me.

(I can't yet view your attachment and did not understand the setup for the second problem.)

3. Oct 23, 2008

### burg25

Would you be able to see the figure if I emailed the figure to you? I do not know how to start the second, so I'm kinda lost there.

4. Oct 23, 2008

### alphysicist

It's been approved now so I can see the attachment. (I'm glad I did not say anything about it; my mental image was way off.)

I think you're doing a few things wrong. In you force equation for block B, you are mixing horizontal and vertical forces (for example, you have the vertical weight and the horizontal friction in the same equation).

You apply $\sum F_x = ma_x$ in one particular direction (when you are dealing with components), so you'll need two equations for block B: one for the vertical force (components) and one for the horizontal forces (components). There are a total of four forces on block B, so what two equations would you get from the force diagram?

The other force equation that you found for the tensions (Tbsin= Tacos$\theta$) look right to me.

What do you get for the weight of block A?

5. Oct 23, 2008

### burg25

Ok so I should sum the forces in the x and y directions. This would give me the equations
Fx= Tb - $$\mu$$mg = 0 for the x since the system is stationary and the y equation would be Fy= Fg=FN. So then I could rewrite the x to be Tb = $$\mu$$mg. So then I would plug this value in for Tb in the equation
Tbsin$$\theta$$ = Tacos$$\theta$$ so
Ta = Tbsin$$\theta$$ / cos$$\theta$$
Ta = (.25(711))sin30/cos30
Ta = 102.6 N
So the maximum mass block A could be to keep the system stationary is 102.6N.
Is this right or did I still screw something up and thank you so much for your help.

6. Oct 23, 2008

### alphysicist

That looks like the right answer to me.

7. Oct 23, 2008

### burg25

Thank you alphysicist.

8. Oct 23, 2008