A large uniform \"butcher block\" rests on two supports and has a weight hanging

Question

A large uniform "butcher block" rests on two supports and has a weight hanging from its end. The block has a mass of 100 kg and a length of 2 meters.

(a) If L = 1.79 meters and the hanging weight is 181 newtons, what is the force on the left support to the nearest newton?

(b) What is the force on the right support?

(c) If the hanging weight is 85 newtons, what is the minimum value for L for the configuration to remain stable to the nearest hundredth of a meter?

(d) What is the maximum value for the hanging weight to the nearest tenth of a kilogram if L = 1.38 meters and the configuration is to remain stable?

Explanation / Answer

a)

Let the force from the right side support be N2 and from the left side support be N1

So, equating torques about the right side support,

N1*L + M*g*(L-1)= 181*(2-L) <------ here (L-1) is the distance of the center of mass of block(M) from the right side support

So, N1 = (181*(2-L) - Mg*(L-1))/L

=(181*(2-1.79) - 100*9.8*(1.79-1))/1.79

= -411.3 N <-----answer ( -ve sign indicates direction towards upwards)

b)

Equating forces in vertical,

N1+181+100*9.8 = N2

So, N2 = 181+100*9.8 + N1

= 181+100*9.8 + (-411.3)

= 749.7 N <-------answer(upwards)

c)

Now, if hanging weight = 85 N

equating the torques about the left end,

N2*L = M*g*1 + 85*2

So, N2*L = (100*9.8+85*2)

So, N2 = (100*9.8+85*2) /L

equating torques about the right support,

N1*L + M*g*(L-1) = 85*(2-L)

So, N1 = (85*(2-L) - 100*9.8*(L-1))/L

Now, equating forces, in vertical

N1 + N2 = 85+Mg =85+100*9.8 = 1065 N

So, (85*(2-L) - 100*9.8*(L-1))/L + (100*9.8+85*2)/L = 85+100*9.8 <----------- (1)

Solving, we get

So, L = 1.08 m <--------answer

d)

From the equation (1) above :

(85*(2-L) - 100*9.8*(L-1))/L + (100*9.8+85*2)/L = 85+100*9.8

Substituting W for 85 and taking L = 1.38 , rewriting the equation we get:

(W*(2-1.38) - 100*9.8*(1.38-1))/1.38 + (100*9.8+W*2)/1.38 = W+100*9.8

So, now, W = 600.6 N <--------answer

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