Question

At what a goal should a 60 kg pager place as ladder against the wall

Answer 1

23. 63 degrees

Step-by-step explanation:

Let us draw a free body diagram.

The net forces on ladder + painter must equal zero. Therefore, along the y-direction:

`N_{}-m_lg-m_p_{}g=0`
`\Rightarrow N=(m_l+m_p)g`

Along the x-direction

`F_r-F_1=0`

Now, we calculate the torque about the about O.

The torque is must be zero:

`-m_pgd\cos _{}\theta-m_lg((l)/(2))\cos \theta+N\cos \theta+F_r\sin \theta=0`

simplifying the above gives

`F_r\sin \theta=(m_pd+m_l((l)/(2))-(N)/(g))g\cos \theta`

Since

`F_r=\mu N=\mu(m_l+m_p)g`

the above becomes

`\mu(m_l+m_p)g\sin \theta=(m_pd+m_l((l)/(2))-(N)/(g))g\cos \theta`
`\Rightarrow\mu(m_l+m_p)\sin \theta=(m_pd+m_l((l)/(2))-(N)/(g))\cos \theta`

dividing both sides by cosine gives

`\mu(m_l+m_p)(\sin\theta)/(\cos\theta)=(m_pd+m_l((l)/(2))-(N)/(g))`

further division gives

`(\sin\theta)/(\cos\theta)=(m_pd+m_l((l)/(2))-(N)/(g))\cdot(1)/(\mu(m_l+m_p))`

substitute the value of N and we get:

`(\sin\theta)/(\cos\theta)=(m_pd+m_l((l)/(2))-((m_l+m_p)g)/(g))\cdot(1)/(\mu(m_l+m_p))`
`(\sin\theta)/(\cos\theta)=(m_pd+m_l((l)/(2))-(m_l+m_p))\cdot(1)/(\mu(m_l+m_p))`

putting in m_p = 60 kg, m_l = 10 kg, l = 6.0 m and d = 1/3 * 6 = 2, and u = 0.5 gives,

`(\sin\theta)/(\cos\theta)=(60\cdot2+10((6)/(2))-(10+60))\cdot(1)/(0.5(10+60))`
`\Rightarrow(\sin \theta)/(\cos \theta)=(16)/(7)`
`\Rightarrow\tan \theta=(16)/(7)`

taking the inverse tan of both sides gives

`\theta=\tan ^(-1)((16)/(7))`
`\theta=66.37^o`

Therefore, the required angle against the wall is

`\alpha=90-\theta=23.63^o`

Hence, the ladder should lean 23.63 degrees against the wall.