Question

What minimum force is required to accelerate a 123 kg wood crate at 0.775 m/s2 up a 22.0 deg concrete slope? u_k = 0.352

Answer 1

Given:

123 kg wood crate is accelerated at 0.775 m/s² up a 22.0° concrete slope.[ u_k = 0.352 ]

To find:

Minimum force required?

Calculation:

• When the object is being accelerated upwards on the inclined plane, both $\sin(\theta)$ component of weight and friction acts downwards.

Let minimum force be F :

$\sf F - ( mg \sin( \theta) + f) = ma$

$\sf\implies F - \{mg \sin( \theta) + \mu_{k}(N) \} = ma$

$\sf \implies F - mg \sin( \theta) - \mu_{k} \{mg \cos( \theta) \} = ma$

$\sf \implies F - mg \{ \sin( \theta) + \mu_{k} \cos( \theta) \} = ma$

$\sf \implies F - mg \{ \sin( {22}^{ \circ} ) + \mu_{k} \cos( {22}^{ \circ} ) \} = ma$

$\sf \implies F - 1230 \{ \sin( {22}^{ \circ} ) + 0.352 \cos( {22}^{ \circ} ) \} = 123(0.775)$

$\sf \implies F = 1230 \{ \sin( {22}^{ \circ} ) + 0.352 \cos( {22}^{ \circ} ) \} + 123(0.775)$

$\sf \implies F = 1230(0.69) + 123(0.775)$

$\sf \implies F = 941.7 \: N$

So, force needed is 941.7 N