A 1.2 kg hammer slams down on a nail at 5.0 m/s and bounces off at 1.0 m/s. If the impact lasts 1.0 ms, what average force is ex
1 answer:
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Answer:
(a). The tension in the cable is 658.33 N.
(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.
(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.
Explanation:
Given that,
Weight of beam= 190 N
Here, The center of gravity is at its center
According to figure,
The angle is
The horizontal component is
The vertical component is
(a). We need to calculate the tension in the cable
Using formula of net torque acting on the pivot
Put the value into the formula
(b). We need to calculate the horizontal components of the force exerted on the beam at the wall
Using formula of horizontal component
Put the value into the formula
(c). We need to calculate the vertical components of the force exerted on the beam at the wall
Using formula of vertical component
Put the value into the formula
Hence, (a). The tension in the cable is 658.33 N.
(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.
(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.
