A uniform crate of mass 240 kg is to be pushed up a loading ramp. Assume the wei

Question

A uniform crate of mass 240 kg is to be pushed up a loading ramp. Assume the weight acts at the geometric centre of the crate. Determine:

(a)  The value of force P to prevent the crate OVERTURNING down the ramp.

(b)  The value of force P to prevent the crate SLIDING down the ramp.

(c)  The minimum value for P necessary to start the transformer sliding up the slope.

(d)  The minimum value for P necessary to overturn the crate UP the ramp.

Assume the coefficient of friction is 0.28 and g = 9.82 m/s2.

A uniform crate of mass 240 kg is to be pushed up a loading ramp. Assume the weight acts at the geometric centre of the crate. Determine: The value of force P to prevent the crate OVERTURNING down the ramp. The value of force P to prevent the crate SLIDING down the ramp. The minimum value for P necessary to start the transformer sliding up the slope. The minimum value for P necessary to overturn the crate UP the ramp. Assume the coefficient of friction is 0.28 and g = 9.82 m/s2. Note, The ramp incline is 25 degrees The crate is 3000mm height and 1000mm wide Contact poistion (were force will be applied) on crate is 2000mm off incline.

Explanation / Answer

a) moment du to P and frictional force = topling force due to weight

P*2+0.28mg cos25 * 1.5 = mg cos25 * 1.5

so 2P = 0.72*1.5*mg*cos25 = 2290.8

or P = 2290.8/2 = 1145.4N

b)frictional force + P = downwards force (mg*sin25)

0.28*mg*cos 25 + P = mg*sin25

0.28*240*9.82*cos25 + P = 240*9.82* sin25

593.9 + P = 989.85

so p = 989.85 - 593.9 = 395.95N

c)

minimum value of P= frictional force + mg*sin 25 = 0.28*240*9.82*cos25 + 240*9.82*sin25 = 593.9+989.85 = 1583.75N

d)

P*2 = 0.28mg*cos25 * 1.5 + mg*cos25*0.5 + mg*sin25 * 1.5 = 3436.2144

so P = 3436.2144/2 = 1718.1P

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